{"id":790,"date":"2019-03-07T13:54:03","date_gmt":"2019-03-07T13:54:03","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/chapter\/solving-trigonometric-equations\/"},"modified":"2020-02-28T15:33:38","modified_gmt":"2020-02-28T15:33:38","slug":"solving-trigonometric-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/chapter\/solving-trigonometric-equations\/","title":{"raw":"3.8 Solving Trigonometric Equations","rendered":"3.8 Solving Trigonometric Equations"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\nIn this section, you will:\r\n<ul>\r\n \t<li>Solve linear trigonometric equations in sine and cosine.<\/li>\r\n \t<li>Solve equations involving a single trigonometric function.<\/li>\r\n \t<li>Solve trigonometric equations using a calculator.<\/li>\r\n \t<li>Solve trigonometric equations that are quadratic in form.<\/li>\r\n \t<li>Solve trigonometric equations using fundamental identities.<\/li>\r\n \t<li>Solve trigonometric equations with multiple angles.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"Figure_07_05_001\" class=\"wp-caption aligncenter\" style=\"width: 941px\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"512\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07135351\/CNX_Precalc_Figure_07_05_001.jpg\" alt=\"Photo of the Egyptian pyramids near a modern city.\" width=\"512\" height=\"163\" \/> Egyptian pyramids standing near a modern city. (credit: Oisin Mulvihill)[\/caption]\r\n\r\n<\/div>\r\n<p id=\"fs-id2261754\">Thales of Miletus (circa 625\u2013547 BC) is known as the founder of geometry. The legend is that he calculated the height of the Great Pyramid of Giza in Egypt using the theory of <em>similar triangles<\/em>, which he developed by measuring the shadow of his staff. Based on proportions, this theory has applications in a number of areas, including fractal geometry, engineering, and architecture. Often, the angle of elevation and the angle of depression are found using similar triangles.<\/p>\r\n<p id=\"fs-id2612701\">In earlier sections of this chapter, we looked at trigonometric identities. Identities are true for all values in the domain of the variable. In this section, we begin our study of trigonometric equations to study real-world scenarios such as the finding the dimensions of the pyramids.<\/p>\r\n\r\n<div id=\"fs-id2123717\" class=\"bc-section section\">\r\n<h3>Solving Linear Trigonometric Equations in Sine and Cosine<\/h3>\r\n<p id=\"fs-id2123619\">Trigonometric equations are, as the name implies, equations that involve trigonometric functions. Similar in many ways to solving polynomial equations or rational equations, only specific values of the variable will be solutions, if there are solutions at all.\u00a0 Additionally, like rational equations, the domain of the function must be considered before we assume that any solution is valid.<\/p>\r\nOften we will solve a trigonometric equation over a specified interval. However, just as often, we will be asked to find all possible solutions, and as trigonometric functions are periodic, solutions are repeated within each period. In other words, trigonometric equations may have an infinite number of solutions.\u00a0 The <span class=\"no-emphasis\">period<\/span> of both the sine function and the cosine function is [latex]2\\pi.[\/latex] In other words, every [latex]2\\pi[\/latex] units, the <em>y-<\/em>values repeat, so\u00a0 [latex]\\mathrm{sin}\\left(\\theta\\right)=\\mathrm{sin}\\left(\\theta \\pm2k\\pi\\right)[\/latex].\u00a0 If we need to find all possible solutions, then we must add [latex]2\\pi k,[\/latex] where [latex]k[\/latex] is an integer, to the initial solution.\r\n<p id=\"fs-id1783046\">There are similar rules for indicating all possible solutions for the other trigonometric functions. Solving trigonometric equations requires the same techniques as solving algebraic equations. We read the equation from left to right, horizontally, like a sentence. We look for known patterns, factor, find common denominators, and substitute certain expressions with a variable to make solving a more straightforward process. However, with trigonometric equations, we also have the advantage of using the identities we developed in the previous sections.<\/p>\r\n\r\n<div id=\"Example_07_05_01\" class=\"textbox examples\">\r\n<div id=\"fs-id2059781\">\r\n<div id=\"fs-id2103256\">\r\n<h3>Example 1:\u00a0 Solving a Linear Trigonometric Equation Involving the Cosine Function<\/h3>\r\n<p id=\"fs-id1618310\">Find all possible exact solutions for the equation [latex]\\mathrm{cos}\\left(\\theta\\right) =\\frac{1}{2}.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1754047\">[reveal-answer q=\"fs-id1754047\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1754047\"]\r\n<p id=\"fs-id1130359\">From the <span class=\"no-emphasis\">unit circle<\/span>, we know that there will be two angles where\u00a0[latex]\\mathrm{cos}\\left(\\theta\\right) =\\frac{1}{2}[\/latex] in one complete revolution, i.e. [latex]0\\le\\theta\\le{2\\pi}.[\/latex] They will occur in the first and fourth quadrants. We recognize [latex]\\frac{1}{2}[\/latex] as a value from one of our special right triangles.\u00a0 We can identify the acute angle as [latex]\\frac{\\pi}{3}.[\/latex]\u00a0 We can then use this as the reference angle to find the angle in the fourth quadrant by computing [latex]2\\pi-\\frac{\\pi}{3}.[\/latex]<\/p>\r\n\r\n<div id=\"fs-id2168355\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}\\mathrm{cos}\\left(\\theta\\right)&amp;=\\frac{1}{2}\\\\ \\theta&amp;=\\frac{\\pi }{3},\\frac{5\\pi }{3}\\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id2823310\">These are the solutions in the interval [latex]\\left[0,2\\pi \\right].[\/latex] All possible solutions are given by\u00a0\u00a0[latex]\\frac{\\pi }{3}\\pm2k\\pi[\/latex] and [latex]\\frac{5\\pi }{3}\\pm2k\\pi [\/latex]\u00a0where [latex]k[\/latex] is an integer.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"Example_07_05_02\" class=\"textbox examples\">\r\n<div id=\"fs-id2449058\">\r\n<div id=\"fs-id1327858\">\r\n<h3>Example 2:\u00a0 Solving a Linear Equation Involving the Sine Function<\/h3>\r\n<p id=\"fs-id1984971\">Find all possible exact solutions for the equation [latex]\\mathrm{sin}\\left(t\\right)=\\frac{1}{2}.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1617496\">[reveal-answer q=\"fs-id1617496\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1617496\"]\r\n<p id=\"fs-id2109893\">Solving for all possible values of <em>t<\/em> means that solutions include angles beyond the period of [latex]2\\pi .[\/latex] From previous work with the unit circle, we know that there are two solutions in one revolution.\u00a0 Since the value is positive, we know these solutions are in the first and second quadrants.\u00a0 From our special right triangles, we know that the angle in the first quadrant is [latex]\\frac{\\pi }{6}[\/latex] and using this as the reference angle, the solution in the second quadrant is [latex]\\frac{5\\pi }{6}.[\/latex]\u00a0 But the problem is asking for all possible values that solve the equation.<\/p>\r\n&nbsp;\r\n\r\nTherefore, the answer is\u00a0\u00a0[latex]\\frac{\\pi }{6}\\pm2\\pi k[\/latex] and [latex]\\frac{5\\pi }{6}\\pm2\\pi k[\/latex]\u00a0\u00a0where [latex]k[\/latex] is an integer.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"Example_07_05_03\" class=\"textbox examples\">\r\n<div id=\"fs-id1936439\">\r\n<div id=\"fs-id1419992\">\r\n<h3>Example 3:\u00a0 Solve the Trigonometric Equation in Linear Form<\/h3>\r\n<p id=\"fs-id2754158\">Solve the equation exactly: [latex]2\\text{ }\\mathrm{cos}\\left(\\theta\\right) -3=-5,\\text{ }\\text{ }0\\le \\theta &lt;2\\pi .[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id2023140\">[reveal-answer q=\"fs-id2023140\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id2023140\"]\r\n<p id=\"fs-id2617859\">Use algebraic techniques to solve the equation.<\/p>\r\n\r\n<div id=\"fs-id1350790\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}2\\text{ }\\mathrm{cos}\\left(\\theta\\right) -3&amp;=-5\\\\ 2\\mathrm{cos}\\left(\\theta\\right) &amp;=-2 &amp;&amp;\\text{Added 3 to both sides. }\\\\ \\mathrm{cos}\\left(\\theta\\right) &amp;=-1&amp;&amp;\\text{Divided both sides by 2. }\\\\ \\theta &amp;=\\pi\\end{align*}[\/latex]<\/div>\r\n<div><\/div>\r\n<div class=\"unnumbered\" style=\"text-align: left\">Thinking of the unit circle, we can see that there is only one place where the cosine value is -1 in one complete revolution.[\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1332935\" class=\"precalculus tryit\">\r\n<h3>Try it #1<\/h3>\r\n<div id=\"ti_07_05_01\">\r\n<div id=\"fs-id1440965\">\r\n<p id=\"fs-id1936123\">Solve exactly the following linear equation on the interval [latex]\\left[0,2\\pi \\right):\\text{ }2\\text{ }\\mathrm{sin}\\left(x\\right)+1=0.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1691784\">[reveal-answer q=\"fs-id1691784\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1691784\"]\r\n<p id=\"fs-id1597318\">[latex]x=\\frac{7\\pi }{6},\\text{ }\\frac{11\\pi }{6}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id2257766\" class=\"bc-section section\">\r\n<h3>Solving Equations Involving a Single Trigonometric Function<\/h3>\r\n<p id=\"fs-id2244514\">When we are given equations that involve only one of the six trigonometric functions, their solutions involve using algebraic techniques and information we know from the unit circle. We need to make several considerations when the equation involves trigonometric functions other than sine and cosine. Problems involving the reciprocals of the primary trigonometric functions need to be viewed from an algebraic perspective. In other words, we will write the equation in terms of the reciprocal function, and solve for the angles using the functions we are most familiar with. Also, an equation involving the tangent function is slightly different from one containing a sine or cosine function. First, as we know, the period of tangent is [latex]\\pi,[\/latex] not [latex]2\\pi.[\/latex] Further, the domain of tangent is all real numbers with the exception of odd integer multiples of [latex]\\frac{\\pi }{2},[\/latex] unless, of course, a problem places its own restrictions on the domain.<\/p>\r\n\r\n<div id=\"Example_07_05_05\" class=\"textbox examples\">\r\n<div id=\"fs-id1161963\">\r\n<div id=\"fs-id1267566\">\r\n<h3>Example 4:\u00a0 Solving a Trigonometric Equation Involving Cosecant<\/h3>\r\n<p id=\"fs-id1256970\">Solve the following equation exactly: [latex]\\mathrm{csc}\\left(\\theta\\right) =-2,\\text{ }0\\le \\theta &lt;4\\pi .[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1056498\">[reveal-answer q=\"fs-id1056498\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1056498\"]\r\n<p id=\"fs-id1114095\">We want all values of [latex]\\theta [\/latex] for which [latex]\\mathrm{csc}\\left(\\theta\\right) =-2[\/latex] over the interval [latex]0\\le \\theta &lt;4\\pi.[\/latex]<\/p>\r\n\r\n<div id=\"fs-id730511\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}\\mathrm{csc}\\left(\\theta\\right) &amp;=-2 \\\\ \\frac{1}{\\mathrm{sin}\\left(\\theta\\right) }&amp;=-2 \\\\ \\mathrm{sin}\\left(\\theta\\right) &amp;=-\\frac{1}{2}\\\\&amp;=\\frac{7\\pi }{6},\\text{ }\\frac{11\\pi }{6},\\text{ }\\frac{19\\pi }{6},\\text{ }\\frac{23\\pi }{6} \\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\r\n<div><\/div>\r\n<div id=\"fs-id1246169\">\r\n<div id=\"fs-id1056498\">\r\n<div>As [latex]\\mathrm{sin}\\left(\\theta\\right) =-\\frac{1}{2},[\/latex] notice that all four solutions are in the third and fourth quadrants.<\/div>\r\n<div>[\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"Example_07_05_06\" class=\"textbox examples\">\r\n<div id=\"fs-id1495254\">\r\n<div id=\"fs-id1304871\">\r\n<h3>Example 5:\u00a0 Solving an Equation Involving Tangent<\/h3>\r\n<p id=\"fs-id2252191\">Solve the equation exactly: [latex]\\mathrm{tan}\\left(\\theta -\\frac{\\pi }{2}\\right)=1,\\text{ } 0\\le \\theta &lt;2\\pi .[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id2239854\">[reveal-answer q=\"fs-id2239854\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id2239854\"]\r\n<p id=\"fs-id1326580\">Recall that the tangent function has a period of [latex]\\pi.[\/latex] On the interval [latex]\\left[0,\\pi \\right),[\/latex] and at the angle of [latex]\\frac{\\pi }{4},[\/latex] the tangent has a value of 1. However, the angle we want is [latex]\\left(\\theta -\\frac{\\pi }{2}\\right).[\/latex] Thus, if [latex]\\mathrm{tan}\\left(\\frac{\\pi }{4}\\right)=1,[\/latex] then<\/p>\r\n\r\n<div id=\"fs-id2141246\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}\\theta -\\frac{\\pi }{2}&amp;=\\frac{\\pi }{4}\\\\ \\theta &amp;=\\frac{3\\pi }{4}\\pm k\\pi \\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id1079211\">Over the interval [latex]\\left[0,2\\pi \\right),[\/latex] we have two solutions:<\/p>\r\n\r\n<div id=\"fs-id3684964\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\frac{3\\pi }{4}[\/latex] and [latex]\\frac{3\\pi}{4}+\\pi =\\frac{7\\pi}{4}[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id761316\" class=\"precalculus tryit\">\r\n<h3>Try it #2<\/h3>\r\n<div id=\"ti_07_05_02\">\r\n<div id=\"fs-id1337984\">\r\n<p id=\"fs-id2886407\">Find all solutions for [latex]\\mathrm{tan}\\left(x\\right)=\\sqrt[\\leftroot{1}\\uproot{2} ]{3}.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1376410\">[reveal-answer q=\"fs-id1376410\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1376410\"]\r\n<p id=\"fs-id1232529\">[latex]\\frac{\\pi }{3}\\pm\\pi k[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"Example_07_05_07\" class=\"textbox examples\">\r\n<div id=\"fs-id1784033\">\r\n<div id=\"fs-id2464356\">\r\n<h3>Example 6:\u00a0 Identify all Solutions to the Equation Involving Tangent<\/h3>\r\n<p id=\"fs-id2031213\">Identify all exact solutions to the equation [latex]2\\left(\\mathrm{tan}\\left(x\\right)+3\\right)=5+\\mathrm{tan}\\left(x\\right),\\text{ }0\\le x&lt;2\\pi .[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1916214\">[reveal-answer q=\"fs-id1916214\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1916214\"]\r\n<p id=\"fs-id708547\">We can solve this equation using only algebra. Isolate the expression [latex]\\mathrm{tan}\\left(x\\right)[\/latex] on the left side of the equals sign.<\/p>\r\n\r\n<div id=\"fs-id2587807\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}2\\text{ }\\left(\\mathrm{tan}\\left(x\\right)\\right)+2\\text{ }\\left(3\\right)&amp; =5+\\mathrm{tan}\\left(x\\right)&amp;&amp;\\text{Distribute the 2 on the left hand side. }\\\\ 2\\text{ }\\mathrm{tan}\\left(x\\right)+6&amp;=5+\\mathrm{tan}\\left(x\\right) \\\\ 2\\text{ }\\mathrm{tan}\\left(x\\right)-\\mathrm{tan}\\left(x\\right)&amp; =5-6 &amp;&amp;\\text{Isolate the tangent on one side.}\\\\ \\mathrm{tan}\\left(x\\right)&amp; =-1 \\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id1517904\">There are two angles on the unit circle that have a tangent value of [latex]-1:\\theta =\\frac{3\\pi }{4}[\/latex] and [latex]\\theta =\\frac{7\\pi }{4}.[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id2352803\" class=\"bc-section section\">\r\n<h3>Solve Trigonometric Equations Using a Calculator<\/h3>\r\n<p id=\"fs-id1925665\">Not all functions can be solved exactly using only the unit circle. When we must solve an equation involving an angle other than one of the special angles, we will need to use a calculator. Make sure it is set to the proper mode, either degrees or radians, depending on the criteria of the given problem.<\/p>\r\n\r\n<div id=\"Example_07_05_08\" class=\"textbox examples\">\r\n<div id=\"fs-id1366869\">\r\n<div id=\"fs-id1805763\">\r\n<h3>Example 7:\u00a0 Using a Calculator to Solve a Trigonometric Equation Involving Sine<\/h3>\r\n<p id=\"fs-id2764602\">Use a calculator to solve the equation [latex]\\mathrm{sin}\\left(\\theta\\right) =0.8,[\/latex] where [latex]\\theta[\/latex] is in radians.<\/p>\r\n\r\n<div id=\"fs-id1133260\">[reveal-answer q=\"fs-id1133260\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1133260\"]\r\n<p id=\"fs-id2644555\">Make sure mode is set to radians. To find [latex]\\theta,[\/latex] use the inverse sine function. On most calculators, you will need to push the 2<sup>ND<\/sup> button and then the SIN button to bring up the [latex]\\mathrm{sin}^{-1}[\/latex] function. What is shown on the screen is\u00a0[latex]\\mathrm{sin}^{-1}\\left(\\text{ }\\right).[\/latex]\u00a0The calculator is ready for the input within the parentheses. For this problem, we enter [latex]\\mathrm{sin}^{-1}\\left(0.8\\right),[\/latex] and press ENTER. Thus, to four decimals places,<\/p>\r\n\r\n<div id=\"fs-id2031037\" class=\"unnumbered\" style=\"text-align: center\">[latex]{\\mathrm{sin}}^{-1}\\left(0.8\\right)\\approx 0.9273[\/latex]<\/div>\r\n<div><\/div>\r\n<p id=\"fs-id1112311\">The solution is [latex]0.9273\\pm2\\pi k.[\/latex]<\/p>\r\nKeep in mind that\u00a0a calculator will only return an angle in quadrants I or IV for the sine function, since that is the range of the inverse sine.\u00a0 Since the sine value is positive, there will be another solution in quadrant 2.\u00a0 The other angle is obtained by using [latex]\\pi -\\theta.[\/latex]\r\n\r\nThis gives us a second set of values in the form\u00a0[latex]2.2143\\pm2\\pi k.[\/latex]\r\n<p id=\"fs-id2780464\">The angle measurements in degrees are based on the first revolution angles of<\/p>\r\n\r\n<div id=\"fs-id2067729\" class=\"unnumbered\">\r\n<p style=\"text-align: center\">[latex]\\theta\\approx {53.1}^{\\circ}\\text{ or } \\theta\\approx {180}^{\\circ}-{53.1}^{\\circ}\\approx {126.9}^{\\circ} [\/latex]<\/p>\r\n\r\n<div id=\"fs-id1354896\">\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<div id=\"fs-id2105271\">\r\n<div id=\"fs-id1622846\">\r\n<h3>Example 8:\u00a0 Using a Calculator to Solve a Trigonometric Equation Involving Secant<\/h3>\r\n<p id=\"fs-id1359705\">Use a calculator to solve the equation [latex]\\mathrm{sec}\\left(\\theta\\right) =-4,[\/latex] giving your answer in radians.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1862382\">[reveal-answer q=\"fs-id1862382\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1862382\"]\r\n<p id=\"fs-id1625296\">We can begin with some algebra.<\/p>\r\n\r\n<div id=\"fs-id2713214\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}\\mathrm{sec}\\left(\\theta\\right) &amp;=-4\\\\ \\frac{1}{\\mathrm{cos}\\left(\\theta\\right)}&amp;=-4\\\\\\mathrm{cos}\\left(\\theta\\right) &amp;=-\\frac{1}{4}\\end{align*}[\/latex]<\/div>\r\n<p id=\"fs-id2176861\">Check that the MODE is in radians. Now use the inverse cosine function.<\/p>\r\n\r\n<div id=\"fs-id1350281\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}\\mathrm{cos}^{-1}\\left(-\\frac{1}{4}\\right)&amp;\\approx 1.8235\\\\ \\theta&amp; \\approx 1.8235+2\\pi k\\\\ \\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1471997\">Since [latex]\\frac{\\pi }{2}\\approx 1.57[\/latex] and [latex]\\pi \\approx 3.14,[\/latex] we know that 1.8235 is between these two numbers, thus [latex]\\theta \\approx \\text{1}\\text{.8235}[\/latex] is in quadrant II. Cosine is also negative in quadrant III. Note that a calculator will only return an angle in quadrants I or II for the cosine function, since that is the range of the inverse cosine. See <a class=\"autogenerated-content\" href=\"#Figure_07_05_005\">Figure 1<\/a>.<\/p>\r\n\r\n<div id=\"Figure_07_05_005\" class=\"medium\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"370\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07135354\/CNX_Precalc_Figure_07_05_005.jpg\" alt=\"Graph of angles theta =approx 1.8235, theta prime =approx pi - 1.8235 = approx 1.3181, and then theta prime = pi + 1.3181 = approx 4.4597\" width=\"370\" height=\"289\" \/> <strong>Figure 1<\/strong>[\/caption]\r\n\r\n<\/div>\r\n<p id=\"fs-id1631910\">So, we also need to find the measure of the angle in quadrant III. In quadrant III, the reference angle is [latex]\\theta^{\\prime} \\approx \\pi -\\text{1}\\text{.8235}\\approx \\text{1}\\text{.3181}\\text{.}[\/latex] The other solution in quadrant III is [latex]\\pi +\\text{1}\\text{.3181}\\approx \\text{4}\\text{.4597.}[\/latex]<\/p>\r\n<p id=\"fs-id1327393\">The solutions are [latex]1.8235\\pm2\\pi k[\/latex] and [latex]4.4597\\pm2\\pi k.[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id2168685\" class=\"precalculus tryit\">\r\n<h3>Try it #3<\/h3>\r\n<div id=\"ti_07_05_03\">\r\n<div id=\"fs-id1868122\">\r\n<p id=\"fs-id2142937\">Solve [latex]\\mathrm{csc}\\left(\\theta\\right) =3.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1562435\">[reveal-answer q=\"fs-id1562435\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1562435\"]\r\n<p id=\"fs-id1334489\">[latex]\\theta \\approx 0.33984\\pm2\\pi k[\/latex] and [latex]\\theta \\approx 2.80176\\pm2\\pi k[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id2461424\" class=\"bc-section section\">\r\n<h3>Solving Trigonometric Equations in Quadratic Form<\/h3>\r\n<p id=\"fs-id2590665\">Solving a <span class=\"no-emphasis\">quadratic equation<\/span> may be more complicated, but once again, we can use algebra as we would for any quadratic equation. Look at the pattern of the equation. Is there more than one trigonometric function in the equation, or is there only one? Which trigonometric function is squared? If there is only one function represented and one of the terms is squared, think about the standard form of a quadratic. Replace the trigonometric function with a variable such as [latex]x[\/latex] or [latex]u.[\/latex] If substitution makes the equation look like a quadratic equation, then we can use the same methods for solving quadratics to solve the trigonometric equations.<\/p>\r\n\r\n<div class=\"textbox examples\">\r\n<h3>How To<\/h3>\r\n<p id=\"fs-id1335125\"><strong>Given a trigonometric equation, solve using algebra<\/strong>.<\/p>\r\n\r\n<ol id=\"fs-id1753734\" type=\"1\">\r\n \t<li>Look for a pattern that suggests an algebraic property, such as the difference of squares or a factoring opportunity.<\/li>\r\n \t<li>Substitute the trigonometric expression with a single variable, such as [latex]x[\/latex] or [latex]u.[\/latex]<\/li>\r\n \t<li>Solve the equation the same way an algebraic equation would be solved.<\/li>\r\n \t<li>Substitute the trigonometric expression back in for the variable in the resulting expressions.<\/li>\r\n \t<li>Solve for the angle.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_07_05_04\" class=\"textbox examples\">\r\n<div id=\"fs-id1081089\">\r\n<div id=\"fs-id1290420\">\r\n<h3>Example 9:\u00a0 Solving a Trigonometric Equation in Quadratic Form Using the Square Root Property<\/h3>\r\n<p id=\"fs-id1354845\">Solve the problem exactly: [latex]2\\text{ }{\\mathrm{sin}}^{2}\\left(\\theta\\right) -1=0, \\text{ }0\\le \\theta &lt;2\\pi .[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1552358\">[reveal-answer q=\"fs-id1552358\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1552358\"]\r\nAs this problem is not easily factored, we will solve using the square root property. First, we use algebra to isolate [latex]\\mathrm{sin}\\left(\\theta\\right).[\/latex] Then we will find the angles.\r\n<div id=\"fs-id1368172\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}2\\mathrm{sin}^{2}\\left(\\theta\\right) -1&amp;=0\\\\ 2\\mathrm{sin}^{2}\\left(\\theta\\right)&amp;=1&amp;&amp;\\text{Add 1 to both sides.} \\\\ \\mathrm{sin}^{2}\\left(\\theta\\right) &amp;=\\frac{1}{2}&amp;&amp;\\text{Divide both sides by 2.}\\\\ \\sqrt{\\mathrm{sin}^{2}\\left(\\theta\\right)}&amp;=\\pm\\sqrt{\\frac{1}{2}}&amp;&amp;\\text{Consider both positive and negative square root values.} \\\\ \\mathrm{sin}\\left(\\theta\\right) &amp;=\\pm\\frac{1}{\\sqrt{2}}&amp;&amp;\\text{Recognize the value from a special right triangle.}\\\\ \\theta &amp;=\\frac{\\pi }{4},\\text{ }\\frac{3\\pi }{4},\\text{ }\\frac{5\\pi }{4},\\text{ }\\frac{7\\pi }{4}&amp;&amp;\\text{Since we have + and -, we have answers in all 4 quadrants}\\end{align*}[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"Example_07_05_10\" class=\"textbox examples\">\r\n<div id=\"fs-id1335579\">\r\n<div id=\"fs-id1897958\">\r\n<h3>Example 10: Solving a Trigonometric Equation in Quadratic Form Using the Quadratic Equation<\/h3>\r\n<p id=\"fs-id2201479\">Solve the equation exactly: [latex]\\mathrm{cos}^{2}\\left(\\theta\\right)+3\\mathrm{cos}\\left(\\theta\\right)-1=0,\\text{ }0\\le \\theta &lt;2\\pi .[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id761598\">[reveal-answer q=\"fs-id761598\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id761598\"]\r\n<p id=\"fs-id1378461\">We begin by using substitution and replacing [latex]\\mathrm{cos}\\left(\\theta\\right)[\/latex] with [latex]u.[\/latex] It is not necessary to use substitution, but it may make the problem easier to solve visually. Let [latex]\\mathrm{cos}\\left(\\theta\\right) =u.[\/latex] We have<\/p>\r\n\r\n<div id=\"fs-id1875852\" class=\"unnumbered\" style=\"text-align: center\">[latex]{u}^{2}+3u-1=0[\/latex][latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id1356287\">The equation cannot be factored, so we will use the <span class=\"no-emphasis\">quadratic formula<\/span> [latex]u=\\frac{-b\\pm\\sqrt{{b}^{2}-4ac}}{2a},[\/latex] where [latex]a=1,\\text{ }b=3[\/latex] and [latex]c=-1.[\/latex]<\/p>\r\n\r\n<div id=\"fs-id2294578\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*} \\\\ u&amp;=\\frac{-3\\pm\\sqrt{{\\left(3\\right)}^{2}-4\\left(1\\right)\\left(-1\\right)}}{2}\\\\ &amp;=\\frac{-3\\pm\\sqrt{13}}{2}\\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\r\nReplace [latex]u[\/latex] with [latex]\\mathrm{cos}\\left(\\theta\\right),[\/latex] and solve. Thus,\r\n<div id=\"fs-id1408206\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*} \\mathrm{cos}\\left(\\theta\\right)&amp;=\\frac{-3\\pm\\sqrt{13}}{2}\\\\\\theta&amp;=\\mathrm{cos}^{-1}\\left(\\frac{-3+\\sqrt{13}}{2}\\right) \\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id1355539\">Note that only the + sign is used. This is because we get an error when we solve [latex]\\theta =\\mathrm{cos}^{-1}\\left(\\frac{-3-\\sqrt{13}}{2}\\right)[\/latex] on a calculator, since the domain of the inverse cosine function is [latex]\\left[-1,1\\right].[\/latex] Therefore the solution is<\/p>\r\n\r\n<div id=\"fs-id1236551\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}\\mathrm{cos}^{-1}\\left(\\frac{-3+\\sqrt{13}}{2}\\right)\\approx 1.26\\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id1298252\">This terminal side of the angle lies in quadrant I. Since cosine is also positive in quadrant IV, the second solution is<\/p>\r\n\r\n<div id=\"fs-id2160568\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}2\\pi -{\\mathrm{cos}}^{-1}\\left(\\frac{-3+\\sqrt{13}}{2}\\right)\\approx 5.02 \\end{align*}[\/latex][latex]\\\\[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"Example_07_05_11\" class=\"textbox examples\">\r\n<div id=\"fs-id1343020\">\r\n<div id=\"fs-id1334881\">\r\n<h3>Example 11:\u00a0 Solving a Trigonometric Equation in Quadratic Form by Factoring<\/h3>\r\n<p id=\"fs-id1403865\">Solve the equation exactly: [latex]2\\text{ }\\mathrm{sin}^{2}\\left(\\theta\\right) -5\\text{ }\\mathrm{sin}\\left(\\theta\\right)+3=0,\\text{ }0\\le \\theta \\le 2\\pi .[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1218136\">[reveal-answer q=\"fs-id1218136\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1218136\"]\r\nUsing grouping, this quadratic can be factored. Either make the real substitution, [latex]\\mathrm{sin}\\left(\\theta\\right) =u[\/latex] and factor [latex]2u^2-5u+3=0,[\/latex] or imagine it, as we factor:\r\n<div id=\"fs-id2676123\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}\\textrm{Without the substitution }&amp;&amp; &amp;&amp;\\text{With the substitution}\\\\2\\text{ }\\mathrm{sin}^{2}\\left(\\theta\\right) -5\\text{ }\\mathrm{sin}\\left(\\theta\\right) +3&amp;=0&amp;&amp; &amp;2u^2-5u+3=0\\\\ \\left(2\\text{ }\\mathrm{sin}\\left(\\theta\\right)-3\\right)\\left(\\mathrm{sin}\\left(\\theta\\right) -1\\right)&amp;=0 &amp;&amp;&amp;\\left(2u-3\\right)\\left(u-1\\right)\\end{align*}[\/latex][latex]\\text{ }[\/latex][latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id1235701\">Now set each factor equal to zero and solve the two equations [latex]2\\text{ }\\mathrm{sin}\\left(\\theta\\right) -3=0[\/latex] and [latex]\\mathrm{sin}\\left(\\theta\\right) -1=0.[\/latex]<\/p>\r\nFor the first equation,\r\n<div id=\"fs-id1634817\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}2\\text{ }\\mathrm{sin}\\left(\\theta\\right) -3&amp;=0\\\\ 2\\text{ }\\mathrm{sin}\\left(\\theta\\right)&amp; =3&amp;&amp;\\textrm{Add 3 to both sides.}\\\\ \\mathrm{sin}\\left(\\theta\\right)&amp; =\\frac{3}{2}&amp;&amp;\\textrm{Divide both sides by 2.} \\\\ \\mathrm{sin}\\left(\\theta\\right)&amp;\\ne \\frac{3}{2}&amp;&amp;\\frac{3}{2}\\textrm{ is not in the domain of the sine function.}\\end{align*}[\/latex][latex]\\text{ }[\/latex][latex]\\\\[\/latex]<\/div>\r\nThe first equation did not have any solution.\u00a0 To solve the second equation,\r\n<div style=\"text-align: center\">[latex]\\begin{align*}\\mathrm{sin}\\left(\\theta\\right) -1&amp;=0\\\\ \\mathrm{sin}\\left(\\theta\\right)&amp;=1&amp;&amp;\\textrm{Add 1 to both sides.}\\\\\\theta&amp;=\\frac{\\pi}{2}&amp;&amp;\\textrm{Sine equals one only for the quadrantal angle.}\\end{align*}[\/latex][latex]\\text{ }[\/latex][latex]\\\\[\/latex]<\/div>\r\n<div id=\"fs-id1218136\">\r\n\r\nThe only solution for this equation is [latex]\\theta=\\frac{\\pi}{2}.[\/latex]\r\n\r\n&nbsp;\r\n<div id=\"fs-id1220222\">\r\n<h3>Analysis<\/h3>\r\nMake sure to check all solutions on the given domain as some factors have no solution. This is because\u00a0the range of the sine function is [latex]\\left[-1,1\\right].[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<div id=\"fs-id2123698\">\r\n<h3>Example 12: Solving an Equation Using an Identity<\/h3>\r\n<p id=\"fs-id2123703\">Solve the equation exactly using an identity: [latex]3\\text{ }\\mathrm{cos}\\left(\\theta\\right)+3=2\\text{ }{\\mathrm{sin}}^{2}\\left(\\theta\\right) ,\\text{ }0\\le \\theta &lt;2\\pi .[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id2131508\">[reveal-answer q=\"fs-id2131508\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id2131508\"]\r\n<p id=\"fs-id2131510\">If we rewrite the right side, we can write the equation in terms of cosine.\u00a0 Recall that the Pythagorean Identity is [latex]\\mathrm{sin}^2\\left(\\theta\\right)+\\mathrm{cos}^2\\left(\\theta\\right)=1[\/latex] and can be solved for [latex]\\mathrm{sin}^{2}\\left(\\theta\\right)[\/latex] so [latex]\\mathrm{sin}^{2}\\left(\\theta\\right)=1-\\mathrm{cos}^{2}\\left(\\theta\\right).[\/latex]<\/p>\r\n\r\n<div id=\"fs-id3026861\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}3\\mathrm{cos}\\left(\\theta\\right)+3&amp;=2\\mathrm{sin}^{2}\\left(\\theta\\right)\\\\ 3\\mathrm{cos}\\left(\\theta\\right)+3&amp;=2\\left(1-\\mathrm{cos}^{2}\\left(\\theta\\right)\\right)&amp;&amp;\\text{Substitute for }\\mathrm{sin}^{2}\\left(\\theta\\right).\\\\\\end{align*}[\/latex]<\/div>\r\n<div><\/div>\r\n<div>Recognize that we now have a quadratic function in [latex]\\mathrm{cos}\\left(\\theta\\right).[\/latex]\u00a0 Factoring can be used to solve this quadratic equation.\u00a0 Below, the same steps used on the left are demonstrated on the right using the substitution method since the quadratic equation in that form may be easier to work with.\u00a0 For the substitution method, let [latex]u=\\mathrm{cos}\\left(\\theta\\right).[\/latex][latex]\\\\[\/latex]<\/div>\r\n<div><\/div>\r\n<div class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}\\textrm{Without the substitution }&amp;&amp; &amp;\\text{With the substitution}\\\\3\\mathrm{cos}\\left(\\theta\\right)+3&amp;=2-2\\mathrm{cos}^{2}\\left(\\theta\\right)&amp;&amp; 3u+3=2-2u^2\\\\ 2\\mathrm{cos}^{2}\\left(\\theta\\right)+3\\mathrm{cos}\\left(\\theta\\right)+1&amp;=0&amp;&amp;2u^2+3u+1=0\\\\ \\left(2\\mathrm{cos}\\left(\\theta\\right)+1\\right)\\left(\\mathrm{cos}\\left(\\theta\\right)+1\\right)&amp;=0&amp;&amp;\\left(2u+1\\right)\\left(u+1\\right)=0\\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\r\n<div><\/div>\r\n<div>Working with the factors using cosine, set each factor equal to zero and solve.[latex]\\\\[\/latex]<\/div>\r\n<div><\/div>\r\n<div style=\"text-align: center\"><span style=\"font-size: 1rem;text-align: initial\">[latex]\\begin{align*}2\\mathrm{cos}\\left(\\theta\\right)+1&amp;=0 &amp;\\mathrm{cos}\\left(\\theta\\right)+1&amp;=0\\\\ \\mathrm{cos}\\left(\\theta\\right)&amp;=-\\frac{1}{2}&amp; \\mathrm{cos}\\left(\\theta\\right)&amp;=-1\\\\\\theta&amp;=\\frac{2\\pi}{3},\\text{ }\\frac{4\\pi}{3}\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }&amp;\\theta&amp;=\\pi\\end{align*}[\/latex][latex]\\\\[\/latex]<\/span><\/div>\r\n<div class=\"unnumbered\" style=\"text-align: center\">\r\n\r\nAgain, remember that there are two quadrants where the [latex]\\mathrm{cos}\\left(\\theta\\right)=-\\frac{1}{2}.[\/latex]\u00a0 These are quadrants 2 and 3.\u00a0 We can find the acute reference angle of [latex]\\frac{\\pi}{3}[\/latex] and use that to generate the solutions in these quadrants.\r\n<div id=\"fs-id2131508\">\r\n<p id=\"fs-id3483085\">Our solutions are [latex]\\frac{2\\pi}{3},\\text{ }\\frac{4\\pi }{3},\\text{ }\\pi.[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1094265\" class=\"precalculus tryit\">\r\n<h3>Try it #4<\/h3>\r\n<div id=\"ti_07_05_04\">\r\n<div id=\"fs-id1213671\">\r\n<p id=\"fs-id1213672\">Solve [latex]\\mathrm{sin}^{2}\\left(\\theta\\right)=2\\mathrm{cos}\\left(\\theta\\right)+2,\\text{ }0\\le\\theta \\le2\\pi [\/latex] [Hint: Make a substitution to express the equation only in terms of cosine.]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1333795\">[reveal-answer q=\"fs-id1333795\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1333795\"]\r\n<p id=\"fs-id2262028\">[latex]\\mathrm{cos}\\left(\\theta\\right) =-1,\\text{ }\\theta =\\pi [\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1892509\" class=\"precalculus tryit\">\r\n<h3>Try it #5<\/h3>\r\n<div id=\"ti_07_05_05\">\r\n<div id=\"fs-id2737020\">\r\n<p id=\"fs-id1743088\">Solve the quadratic equation [latex]2\\mathrm{cos}^{2}\\left(\\theta\\right) +\\mathrm{cos}\\left(\\theta\\right) =0.[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id3549256\">[reveal-answer q=\"fs-id3549256\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id3549256\"]\r\n<p id=\"fs-id3549257\">[latex]\\frac{\\pi }{2},\\text{ }\\frac{2\\pi }{3},\\text{ }\\frac{4\\pi }{3},\\text{ }\\frac{3\\pi }{2}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1597424\" class=\"bc-section section\">\r\n<h3>Solving Trigonometric Equations Using Fundamental Identities<\/h3>\r\n<p id=\"fs-id1396319\">While algebra can be used to solve a number of trigonometric equations, we can also use the fundamental identities because they make solving equations simpler. Remember that the techniques we use for solving are not the same as those for verifying identities. The basic rules of algebra apply here, as opposed to rewriting one side of the identity to match the other side. In the next example, we use two identities to simplify the equation.<\/p>\r\nWe will need to use some new identities in this section.\u00a0 They are called the\u00a0<strong>double-angle<\/strong> identities.\u00a0We will not take the time to show where the following identities come from. Keep in mind that there are other trigonometric identities that we have not covered in this material. If you are interested, you can look up the sum and difference formulas for sine and cosine, and use those to generate some other identities, including the ones shown below.\r\n<div class=\"textbox definitions\">\r\n<h3>Definition<\/h3>\r\n<p id=\"fs-id1165137661589\">The double angle identities are:<\/p>\r\n<p style=\"text-align: center\">[latex]\\mathrm{sin}\\left(2\\theta\\right)=2\\mathrm{sin}\\left(\\theta\\right)\\mathrm{cos}\\left(\\theta\\right)[\/latex]<\/p>\r\n[latex]\\\\[\/latex]\r\n<p style=\"text-align: center\">[latex]\\begin{align*}\\mathrm{cos}\\left(2\\theta\\right)&amp;=\\mathrm{cos}^{2}\\left(\\theta\\right)-\\mathrm{sin}^{2}\\left(\\theta\\right)\\text{ or,}\\\\&amp;=1-2\\mathrm{sin}^{2}\\left(\\theta\\right)\\text{ or,}\\\\&amp;=2\\mathrm{cos}^{2}\\left(\\theta\\right)-1\\end{align*}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"Example_07_05_15\" class=\"textbox examples\">\r\n<div id=\"fs-id1916902\">\r\n<div id=\"fs-id1916904\">\r\n<h3>Example 13: Solving the Equation Using a Double-Angle Formula<\/h3>\r\n<p id=\"fs-id1916909\">Solve the equation exactly using a double-angle formula: [latex]\\mathrm{cos}\\left(2\\theta\\right)=\\mathrm{cos}\\left(\\theta\\right).[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id697866\">[reveal-answer q=\"fs-id697866\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id697866\"]\r\n<p id=\"fs-id697868\">We have three choices of expressions to substitute for the double-angle of cosine. As it is simpler to solve for one trigonometric function at a time, we will choose the double-angle identity involving only cosine:<\/p>\r\n\r\n<div id=\"fs-id2070377\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}\\mathrm{cos}\\left(2\\theta\\right)&amp;=\\mathrm{cos}\\left(\\theta\\right)\\\\ 2\\mathrm{cos}^{2}\\left(\\theta\\right)-1&amp;=\\mathrm{cos}\\left(\\theta\\right)&amp;&amp;\\text{Replace left hand side with identity.}\\\\ 2\\mathrm{cos}^{2}\\left(\\theta\\right)-\\mathrm{cos}\\left(\\theta\\right)-1&amp;=0&amp;&amp;\\text{Move all terms to one side.}\\\\ \\left(2\\mathrm{cos}\\left(\\theta\\right)+1\\right)\\left(\\mathrm{cos}\\left(\\theta\\right)-1\\right)&amp;=0&amp;&amp;\\text{Factor the left hand side.}\\end{align*}[\/latex]<\/div>\r\n[latex]\\\\[\/latex]\r\n\r\nSet the factors equal to zero and solve.\r\n<p style=\"text-align: center\">[latex]\\begin{align*}2\\mathrm{cos}\\left(\\theta\\right)+1&amp;=0&amp;\\mathrm{cos}\\left(\\theta\\right)-1&amp;=0\\\\\\mathrm{cos}\\left(\\theta\\right)&amp;=-\\frac{1}{2}&amp;\\mathrm{cos}\\left(\\theta\\right)&amp;=1\\\\\\theta&amp;=\\frac{2\\pi}{3}\\pm2\\pi k&amp;\\theta&amp;=0\\pm2\\pi k\\\\\\theta&amp;=\\frac{4\\pi}{3}\\pm2\\pi k\\end{align*}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1518803\" class=\"bc-section section\">\r\n<h3>Solving Trigonometric Equations with Multiple Angles<\/h3>\r\n<p id=\"fs-id1518808\">Sometimes it is not possible to solve a trigonometric equation with identities that have a multiple angle, such as [latex]\\mathrm{sin}\\left(2x\\right)[\/latex] or [latex]\\mathrm{cos}\\left(3x\\right).[\/latex] When confronted with these equations, recall that [latex]y=\\mathrm{sin}\\left(2x\\right)[\/latex] is a <span class=\"no-emphasis\">horizontal compression<\/span> by a factor of 2 of the function [latex]y=\\mathrm{sin}\\left(x\\right).[\/latex]\u00a0 On an interval of [latex]2\\pi,[\/latex] we can graph two periods of [latex]y=\\mathrm{sin}\\left(2x\\right),[\/latex] as opposed to one cycle of [latex]y=\\mathrm{sin}\\left(x\\right).[\/latex] This compression of the graph leads us to believe there may be twice as many <em>x<\/em>-intercepts or solutions to [latex]\\mathrm{sin}\\left(2x\\right)=0[\/latex] compared to [latex]\\mathrm{sin}\\left(x\\right)=0.[\/latex] This information will help us solve the similar type of equation shown in the example.<\/p>\r\n\r\n<div id=\"Example_07_05_17\" class=\"textbox examples\">\r\n<div id=\"fs-id2223187\">\r\n<div id=\"fs-id2223189\">\r\n<h3>Example 14: Solving a Multiple Angle Trigonometric Equation<\/h3>\r\n<p id=\"fs-id2223194\">Solve exactly: [latex]\\mathrm{cos}\\left(2x\\right)=\\frac{1}{2}[\/latex] on [latex]\\left[0,2\\pi \\right).[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id2740923\">[reveal-answer q=\"fs-id2740923\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id2740923\"]\r\n<p id=\"fs-id2740926\">We can see that this equation is the standard equation with a multiple of an angle. If [latex]\\mathrm{cos}\\left(\\theta\\right)=\\frac{1}{2},[\/latex] we know [latex]\\theta[\/latex] is in quadrants I and IV. While [latex]\\theta =\\mathrm{cos}^{-1}\\left(\\frac{1}{2}\\right)[\/latex] will only yield solutions in quadrants I and II because of the range of the inverse cosine function, we recognize that the solutions to the equation [latex]\\mathrm{cos}\\left(\\theta\\right)=\\frac{1}{2}[\/latex] will be in quadrants I and IV using ideas from our unit circle.<\/p>\r\n<p id=\"fs-id3162261\">Therefore, the possible angles are [latex]\\theta=\\frac{\\pi }{3}[\/latex] and [latex]\\theta=\\frac{5\\pi}{3}.[\/latex] So, [latex]2x=\\frac{\\pi}{3}[\/latex] or [latex]2x=\\frac{5\\pi}{3},[\/latex] which means that [latex]x=\\frac{\\pi}{6}[\/latex] or [latex]x=\\frac{5\\pi}{6}.[\/latex] [latex]\\\\[\/latex]Does this make sense? Yes, because [latex]\\mathrm{cos}\\left(2\\left(\\frac{\\pi}{6}\\right)\\right)=\\mathrm{cos}\\left(\\frac{\\pi}{3}\\right)=\\frac{1}{2}.[\/latex]<\/p>\r\n<p id=\"fs-id1966562\">Are there any other possible answers? Let us return to our first step.<\/p>\r\n<p id=\"fs-id1966566\">In quadrant I, [latex]2x=\\frac{\\pi}{3},[\/latex] so [latex]x=\\frac{\\pi}{6}[\/latex] as noted. Let us revolve around the circle again:<\/p>\r\n\r\n<div id=\"fs-id1893406\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}2x&amp;=\\frac{\\pi}{3}+2\\pi \\\\ 2x&amp;=\\frac{\\pi }{3}+\\frac{6\\pi}{3}\\\\ 2x&amp;=\\frac{7\\pi }{3}\\\\x&amp;=\\frac{7\\pi}{6}\\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id1805664\">One more rotation yields<\/p>\r\n\r\n<div id=\"fs-id1805667\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}2x&amp;=\\frac{\\pi }{3}+4\\pi \\\\2x&amp;=\\frac{\\pi}{3}+\\frac{12\\pi}{3}\\\\ 2x&amp;=\\frac{13\\pi}{3}\\\\x&amp;=\\frac{13\\pi}{6}\\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id3654355\">[latex]x=\\frac{13\\pi}{6}&gt;2\\pi,[\/latex] so this value for [latex]x[\/latex] is larger than [latex]2\\pi,[\/latex] so it is not a solution on [latex]\\left[0,2\\pi \\right).[\/latex]<\/p>\r\n<p id=\"fs-id2748825\">In quadrant IV, [latex]2x=\\frac{5\\pi }{3},[\/latex] so [latex]x=\\frac{5\\pi }{6}[\/latex] as noted. Let us revolve around the circle again:<\/p>\r\n\r\n<div id=\"fs-id1154337\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}2x&amp;=\\frac{5\\pi}{3}+2\\pi \\\\2x&amp;=\\frac{5\\pi }{3}+\\frac{6\\pi }{3} \\\\ 2x&amp;=\\frac{11\\pi }{3}\\\\x&amp;=\\frac{11\\pi}{6} \\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id2123818\">One more rotation yields<\/p>\r\n\r\n<div id=\"fs-id2123822\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}2x&amp;=\\frac{5\\pi }{3}+4\\pi \\\\ 2x&amp;=\\frac{5\\pi }{3}+\\frac{12\\pi }{3} \\\\ 2x&amp;=\\frac{17\\pi }{3}\\\\x&amp;=\\frac{17\\pi}{6}\\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\r\n<p id=\"fs-id1217412\">[latex]x=\\frac{17\\pi }{6}&gt;2\\pi ,[\/latex] so this value for [latex]x[\/latex] is larger than [latex]2\\pi,[\/latex] which means it is not a solution on [latex]\\left[0,2\\pi \\right).[\/latex]<\/p>\r\n<p id=\"fs-id1676669\">Our solutions are [latex]\\frac{\\pi}{6},\\text{ }\\frac{5\\pi }{6},\\text{ }\\frac{7\\pi }{6},[\/latex] and [latex]\\frac{11\\pi }{6}.[\/latex] Note that whenever we solve a problem in the form of [latex]\\mathrm{sin}\\left(nx\\right)=c[\/latex] or [latex]\\mathrm{cos}\\left(nx\\right)=c,[\/latex] we must go around the unit circle [latex]{n}[\/latex] times.<\/p>\r\nWe can see this easily if we first write the solution for [latex]2x[\/latex] in the general form.\u00a0 One of these equations is shown below:\r\n<p style=\"text-align: center\">[latex]\\begin{align*}2x&amp;=\\frac{\\pi}{3}+2\\pi\\cdot k\\\\x&amp;=\\frac{\\pi}{6}+\\pi \\cdot k&amp;&amp;\\text{Divide both sides by 2.}\\end{align*}[\/latex][latex]\\\\[\/latex]<\/p>\r\nWe can now see that adding [latex]\\pi[\/latex] to [latex]\\frac{\\pi}{6}[\/latex] gives us\u00a0[latex]\\frac{7\\pi}{6}[\/latex] and that adding [latex]2\\cdot\\pi[\/latex] would give us\u00a0[latex]\\frac{13\\pi}{6}[\/latex]\u00a0 which is larger than [latex]2\\pi.[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id2267036\" class=\"precalculus media\">\r\n<p id=\"fs-id2267042\">Access these online resources for additional instruction and practice with solving trigonometric equations.<\/p>\r\n\r\n<ul id=\"fs-id2267047\">\r\n \t<li><a href=\"http:\/\/openstax.org\/l\/solvetrigeqI\">Solving Trigonometric Equations I<\/a><\/li>\r\n \t<li><a href=\"https:\/\/www.youtube.com\/watch?v=ABKO3ta_Azw\">Solving Trigonometric Equations II<\/a><\/li>\r\n \t<li><a href=\"http:\/\/openstax.org\/l\/solvetrigeqIII\">Solving Trigonometric Equations III<\/a><\/li>\r\n \t<li><a href=\"http:\/\/openstax.org\/l\/solvetrigeqIV\">Solving Trigonometric Equations IV<\/a><\/li>\r\n \t<li><a href=\"http:\/\/openstax.org\/l\/solvetrigeqV\">Solving Trigonometric Equations V<\/a><\/li>\r\n \t<li><a href=\"http:\/\/openstax.org\/l\/solvetrigeqVI\">Solving Trigonometric Equations VI<\/a><\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id2267085\" class=\"textbox key-takeaways\">\r\n<h3>Key Concepts<\/h3>\r\n<ul id=\"fs-id2267091\">\r\n \t<li>When solving linear trigonometric equations, we can use algebraic techniques just as we do solving algebraic equations. Look for patterns, like the difference of squares, quadratic form, or an expression that lends itself well to substitution.<\/li>\r\n \t<li>Equations involving a single trigonometric function can be solved or verified using the unit circle.<\/li>\r\n \t<li>We can also solve trigonometric equations using a graphing calculator.<\/li>\r\n \t<li>Many equations appear quadratic in form. We can use substitution to make the equation appear simpler, and then use the same techniques we use solving an algebraic quadratic: factoring, the quadratic formula, etc.<\/li>\r\n \t<li>We can also use the identities to solve trigonometric equation.<\/li>\r\n \t<li>We can use substitution to solve a multiple-angle trigonometric equation, which is a compression of a standard trigonometric function. We will need to take the compression into account and verify that we have found all solutions on the given interval.<\/li>\r\n<\/ul>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<p>In this section, you will:<\/p>\n<ul>\n<li>Solve linear trigonometric equations in sine and cosine.<\/li>\n<li>Solve equations involving a single trigonometric function.<\/li>\n<li>Solve trigonometric equations using a calculator.<\/li>\n<li>Solve trigonometric equations that are quadratic in form.<\/li>\n<li>Solve trigonometric equations using fundamental identities.<\/li>\n<li>Solve trigonometric equations with multiple angles.<\/li>\n<\/ul>\n<\/div>\n<div id=\"Figure_07_05_001\" class=\"wp-caption aligncenter\" style=\"width: 941px\">\n<div style=\"width: 522px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07135351\/CNX_Precalc_Figure_07_05_001.jpg\" alt=\"Photo of the Egyptian pyramids near a modern city.\" width=\"512\" height=\"163\" \/><\/p>\n<p class=\"wp-caption-text\">Egyptian pyramids standing near a modern city. (credit: Oisin Mulvihill)<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id2261754\">Thales of Miletus (circa 625\u2013547 BC) is known as the founder of geometry. The legend is that he calculated the height of the Great Pyramid of Giza in Egypt using the theory of <em>similar triangles<\/em>, which he developed by measuring the shadow of his staff. Based on proportions, this theory has applications in a number of areas, including fractal geometry, engineering, and architecture. Often, the angle of elevation and the angle of depression are found using similar triangles.<\/p>\n<p id=\"fs-id2612701\">In earlier sections of this chapter, we looked at trigonometric identities. Identities are true for all values in the domain of the variable. In this section, we begin our study of trigonometric equations to study real-world scenarios such as the finding the dimensions of the pyramids.<\/p>\n<div id=\"fs-id2123717\" class=\"bc-section section\">\n<h3>Solving Linear Trigonometric Equations in Sine and Cosine<\/h3>\n<p id=\"fs-id2123619\">Trigonometric equations are, as the name implies, equations that involve trigonometric functions. Similar in many ways to solving polynomial equations or rational equations, only specific values of the variable will be solutions, if there are solutions at all.\u00a0 Additionally, like rational equations, the domain of the function must be considered before we assume that any solution is valid.<\/p>\n<p>Often we will solve a trigonometric equation over a specified interval. However, just as often, we will be asked to find all possible solutions, and as trigonometric functions are periodic, solutions are repeated within each period. In other words, trigonometric equations may have an infinite number of solutions.\u00a0 The <span class=\"no-emphasis\">period<\/span> of both the sine function and the cosine function is [latex]2\\pi.[\/latex] In other words, every [latex]2\\pi[\/latex] units, the <em>y-<\/em>values repeat, so\u00a0 [latex]\\mathrm{sin}\\left(\\theta\\right)=\\mathrm{sin}\\left(\\theta \\pm2k\\pi\\right)[\/latex].\u00a0 If we need to find all possible solutions, then we must add [latex]2\\pi k,[\/latex] where [latex]k[\/latex] is an integer, to the initial solution.<\/p>\n<p id=\"fs-id1783046\">There are similar rules for indicating all possible solutions for the other trigonometric functions. Solving trigonometric equations requires the same techniques as solving algebraic equations. We read the equation from left to right, horizontally, like a sentence. We look for known patterns, factor, find common denominators, and substitute certain expressions with a variable to make solving a more straightforward process. However, with trigonometric equations, we also have the advantage of using the identities we developed in the previous sections.<\/p>\n<div id=\"Example_07_05_01\" class=\"textbox examples\">\n<div id=\"fs-id2059781\">\n<div id=\"fs-id2103256\">\n<h3>Example 1:\u00a0 Solving a Linear Trigonometric Equation Involving the Cosine Function<\/h3>\n<p id=\"fs-id1618310\">Find all possible exact solutions for the equation [latex]\\mathrm{cos}\\left(\\theta\\right) =\\frac{1}{2}.[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1754047\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1754047\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1754047\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1130359\">From the <span class=\"no-emphasis\">unit circle<\/span>, we know that there will be two angles where\u00a0[latex]\\mathrm{cos}\\left(\\theta\\right) =\\frac{1}{2}[\/latex] in one complete revolution, i.e. [latex]0\\le\\theta\\le{2\\pi}.[\/latex] They will occur in the first and fourth quadrants. We recognize [latex]\\frac{1}{2}[\/latex] as a value from one of our special right triangles.\u00a0 We can identify the acute angle as [latex]\\frac{\\pi}{3}.[\/latex]\u00a0 We can then use this as the reference angle to find the angle in the fourth quadrant by computing [latex]2\\pi-\\frac{\\pi}{3}.[\/latex]<\/p>\n<div id=\"fs-id2168355\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}\\mathrm{cos}\\left(\\theta\\right)&=\\frac{1}{2}\\\\ \\theta&=\\frac{\\pi }{3},\\frac{5\\pi }{3}\\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id2823310\">These are the solutions in the interval [latex]\\left[0,2\\pi \\right].[\/latex] All possible solutions are given by\u00a0\u00a0[latex]\\frac{\\pi }{3}\\pm2k\\pi[\/latex] and [latex]\\frac{5\\pi }{3}\\pm2k\\pi[\/latex]\u00a0where [latex]k[\/latex] is an integer.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_07_05_02\" class=\"textbox examples\">\n<div id=\"fs-id2449058\">\n<div id=\"fs-id1327858\">\n<h3>Example 2:\u00a0 Solving a Linear Equation Involving the Sine Function<\/h3>\n<p id=\"fs-id1984971\">Find all possible exact solutions for the equation [latex]\\mathrm{sin}\\left(t\\right)=\\frac{1}{2}.[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1617496\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1617496\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1617496\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id2109893\">Solving for all possible values of <em>t<\/em> means that solutions include angles beyond the period of [latex]2\\pi .[\/latex] From previous work with the unit circle, we know that there are two solutions in one revolution.\u00a0 Since the value is positive, we know these solutions are in the first and second quadrants.\u00a0 From our special right triangles, we know that the angle in the first quadrant is [latex]\\frac{\\pi }{6}[\/latex] and using this as the reference angle, the solution in the second quadrant is [latex]\\frac{5\\pi }{6}.[\/latex]\u00a0 But the problem is asking for all possible values that solve the equation.<\/p>\n<p>&nbsp;<\/p>\n<p>Therefore, the answer is\u00a0\u00a0[latex]\\frac{\\pi }{6}\\pm2\\pi k[\/latex] and [latex]\\frac{5\\pi }{6}\\pm2\\pi k[\/latex]\u00a0\u00a0where [latex]k[\/latex] is an integer.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_07_05_03\" class=\"textbox examples\">\n<div id=\"fs-id1936439\">\n<div id=\"fs-id1419992\">\n<h3>Example 3:\u00a0 Solve the Trigonometric Equation in Linear Form<\/h3>\n<p id=\"fs-id2754158\">Solve the equation exactly: [latex]2\\text{ }\\mathrm{cos}\\left(\\theta\\right) -3=-5,\\text{ }\\text{ }0\\le \\theta <2\\pi .[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id2023140\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id2023140\">Show Solution<\/span><\/p>\n<div id=\"qfs-id2023140\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id2617859\">Use algebraic techniques to solve the equation.<\/p>\n<div id=\"fs-id1350790\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}2\\text{ }\\mathrm{cos}\\left(\\theta\\right) -3&=-5\\\\ 2\\mathrm{cos}\\left(\\theta\\right) &=-2 &&\\text{Added 3 to both sides. }\\\\ \\mathrm{cos}\\left(\\theta\\right) &=-1&&\\text{Divided both sides by 2. }\\\\ \\theta &=\\pi\\end{align*}[\/latex]<\/div>\n<div><\/div>\n<div class=\"unnumbered\" style=\"text-align: left\">Thinking of the unit circle, we can see that there is only one place where the cosine value is -1 in one complete revolution.<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1332935\" class=\"precalculus tryit\">\n<h3>Try it #1<\/h3>\n<div id=\"ti_07_05_01\">\n<div id=\"fs-id1440965\">\n<p id=\"fs-id1936123\">Solve exactly the following linear equation on the interval [latex]\\left[0,2\\pi \\right):\\text{ }2\\text{ }\\mathrm{sin}\\left(x\\right)+1=0.[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1691784\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1691784\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1691784\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1597318\">[latex]x=\\frac{7\\pi }{6},\\text{ }\\frac{11\\pi }{6}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id2257766\" class=\"bc-section section\">\n<h3>Solving Equations Involving a Single Trigonometric Function<\/h3>\n<p id=\"fs-id2244514\">When we are given equations that involve only one of the six trigonometric functions, their solutions involve using algebraic techniques and information we know from the unit circle. We need to make several considerations when the equation involves trigonometric functions other than sine and cosine. Problems involving the reciprocals of the primary trigonometric functions need to be viewed from an algebraic perspective. In other words, we will write the equation in terms of the reciprocal function, and solve for the angles using the functions we are most familiar with. Also, an equation involving the tangent function is slightly different from one containing a sine or cosine function. First, as we know, the period of tangent is [latex]\\pi,[\/latex] not [latex]2\\pi.[\/latex] Further, the domain of tangent is all real numbers with the exception of odd integer multiples of [latex]\\frac{\\pi }{2},[\/latex] unless, of course, a problem places its own restrictions on the domain.<\/p>\n<div id=\"Example_07_05_05\" class=\"textbox examples\">\n<div id=\"fs-id1161963\">\n<div id=\"fs-id1267566\">\n<h3>Example 4:\u00a0 Solving a Trigonometric Equation Involving Cosecant<\/h3>\n<p id=\"fs-id1256970\">Solve the following equation exactly: [latex]\\mathrm{csc}\\left(\\theta\\right) =-2,\\text{ }0\\le \\theta <4\\pi .[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1056498\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1056498\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1056498\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1114095\">We want all values of [latex]\\theta[\/latex] for which [latex]\\mathrm{csc}\\left(\\theta\\right) =-2[\/latex] over the interval [latex]0\\le \\theta <4\\pi.[\/latex]<\/p>\n<div id=\"fs-id730511\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}\\mathrm{csc}\\left(\\theta\\right) &=-2 \\\\ \\frac{1}{\\mathrm{sin}\\left(\\theta\\right) }&=-2 \\\\ \\mathrm{sin}\\left(\\theta\\right) &=-\\frac{1}{2}\\\\&=\\frac{7\\pi }{6},\\text{ }\\frac{11\\pi }{6},\\text{ }\\frac{19\\pi }{6},\\text{ }\\frac{23\\pi }{6} \\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\n<div><\/div>\n<div id=\"fs-id1246169\">\n<div id=\"fs-id1056498\">\n<div>As [latex]\\mathrm{sin}\\left(\\theta\\right) =-\\frac{1}{2},[\/latex] notice that all four solutions are in the third and fourth quadrants.<\/div>\n<div><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_07_05_06\" class=\"textbox examples\">\n<div id=\"fs-id1495254\">\n<div id=\"fs-id1304871\">\n<h3>Example 5:\u00a0 Solving an Equation Involving Tangent<\/h3>\n<p id=\"fs-id2252191\">Solve the equation exactly: [latex]\\mathrm{tan}\\left(\\theta -\\frac{\\pi }{2}\\right)=1,\\text{ } 0\\le \\theta <2\\pi .[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id2239854\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id2239854\">Show Solution<\/span><\/p>\n<div id=\"qfs-id2239854\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1326580\">Recall that the tangent function has a period of [latex]\\pi.[\/latex] On the interval [latex]\\left[0,\\pi \\right),[\/latex] and at the angle of [latex]\\frac{\\pi }{4},[\/latex] the tangent has a value of 1. However, the angle we want is [latex]\\left(\\theta -\\frac{\\pi }{2}\\right).[\/latex] Thus, if [latex]\\mathrm{tan}\\left(\\frac{\\pi }{4}\\right)=1,[\/latex] then<\/p>\n<div id=\"fs-id2141246\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}\\theta -\\frac{\\pi }{2}&=\\frac{\\pi }{4}\\\\ \\theta &=\\frac{3\\pi }{4}\\pm k\\pi \\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id1079211\">Over the interval [latex]\\left[0,2\\pi \\right),[\/latex] we have two solutions:<\/p>\n<div id=\"fs-id3684964\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\frac{3\\pi }{4}[\/latex] and [latex]\\frac{3\\pi}{4}+\\pi =\\frac{7\\pi}{4}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id761316\" class=\"precalculus tryit\">\n<h3>Try it #2<\/h3>\n<div id=\"ti_07_05_02\">\n<div id=\"fs-id1337984\">\n<p id=\"fs-id2886407\">Find all solutions for [latex]\\mathrm{tan}\\left(x\\right)=\\sqrt[\\leftroot{1}\\uproot{2} ]{3}.[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1376410\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1376410\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1376410\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1232529\">[latex]\\frac{\\pi }{3}\\pm\\pi k[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_07_05_07\" class=\"textbox examples\">\n<div id=\"fs-id1784033\">\n<div id=\"fs-id2464356\">\n<h3>Example 6:\u00a0 Identify all Solutions to the Equation Involving Tangent<\/h3>\n<p id=\"fs-id2031213\">Identify all exact solutions to the equation [latex]2\\left(\\mathrm{tan}\\left(x\\right)+3\\right)=5+\\mathrm{tan}\\left(x\\right),\\text{ }0\\le x<2\\pi .[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1916214\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1916214\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1916214\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id708547\">We can solve this equation using only algebra. Isolate the expression [latex]\\mathrm{tan}\\left(x\\right)[\/latex] on the left side of the equals sign.<\/p>\n<div id=\"fs-id2587807\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}2\\text{ }\\left(\\mathrm{tan}\\left(x\\right)\\right)+2\\text{ }\\left(3\\right)& =5+\\mathrm{tan}\\left(x\\right)&&\\text{Distribute the 2 on the left hand side. }\\\\ 2\\text{ }\\mathrm{tan}\\left(x\\right)+6&=5+\\mathrm{tan}\\left(x\\right) \\\\ 2\\text{ }\\mathrm{tan}\\left(x\\right)-\\mathrm{tan}\\left(x\\right)& =5-6 &&\\text{Isolate the tangent on one side.}\\\\ \\mathrm{tan}\\left(x\\right)& =-1 \\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id1517904\">There are two angles on the unit circle that have a tangent value of [latex]-1:\\theta =\\frac{3\\pi }{4}[\/latex] and [latex]\\theta =\\frac{7\\pi }{4}.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id2352803\" class=\"bc-section section\">\n<h3>Solve Trigonometric Equations Using a Calculator<\/h3>\n<p id=\"fs-id1925665\">Not all functions can be solved exactly using only the unit circle. When we must solve an equation involving an angle other than one of the special angles, we will need to use a calculator. Make sure it is set to the proper mode, either degrees or radians, depending on the criteria of the given problem.<\/p>\n<div id=\"Example_07_05_08\" class=\"textbox examples\">\n<div id=\"fs-id1366869\">\n<div id=\"fs-id1805763\">\n<h3>Example 7:\u00a0 Using a Calculator to Solve a Trigonometric Equation Involving Sine<\/h3>\n<p id=\"fs-id2764602\">Use a calculator to solve the equation [latex]\\mathrm{sin}\\left(\\theta\\right) =0.8,[\/latex] where [latex]\\theta[\/latex] is in radians.<\/p>\n<div id=\"fs-id1133260\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1133260\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1133260\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id2644555\">Make sure mode is set to radians. To find [latex]\\theta,[\/latex] use the inverse sine function. On most calculators, you will need to push the 2<sup>ND<\/sup> button and then the SIN button to bring up the [latex]\\mathrm{sin}^{-1}[\/latex] function. What is shown on the screen is\u00a0[latex]\\mathrm{sin}^{-1}\\left(\\text{ }\\right).[\/latex]\u00a0The calculator is ready for the input within the parentheses. For this problem, we enter [latex]\\mathrm{sin}^{-1}\\left(0.8\\right),[\/latex] and press ENTER. Thus, to four decimals places,<\/p>\n<div id=\"fs-id2031037\" class=\"unnumbered\" style=\"text-align: center\">[latex]{\\mathrm{sin}}^{-1}\\left(0.8\\right)\\approx 0.9273[\/latex]<\/div>\n<div><\/div>\n<p id=\"fs-id1112311\">The solution is [latex]0.9273\\pm2\\pi k.[\/latex]<\/p>\n<p>Keep in mind that\u00a0a calculator will only return an angle in quadrants I or IV for the sine function, since that is the range of the inverse sine.\u00a0 Since the sine value is positive, there will be another solution in quadrant 2.\u00a0 The other angle is obtained by using [latex]\\pi -\\theta.[\/latex]<\/p>\n<p>This gives us a second set of values in the form\u00a0[latex]2.2143\\pm2\\pi k.[\/latex]<\/p>\n<p id=\"fs-id2780464\">The angle measurements in degrees are based on the first revolution angles of<\/p>\n<div id=\"fs-id2067729\" class=\"unnumbered\">\n<p style=\"text-align: center\">[latex]\\theta\\approx {53.1}^{\\circ}\\text{ or } \\theta\\approx {180}^{\\circ}-{53.1}^{\\circ}\\approx {126.9}^{\\circ}[\/latex]<\/p>\n<div id=\"fs-id1354896\">\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<div id=\"fs-id2105271\">\n<div id=\"fs-id1622846\">\n<h3>Example 8:\u00a0 Using a Calculator to Solve a Trigonometric Equation Involving Secant<\/h3>\n<p id=\"fs-id1359705\">Use a calculator to solve the equation [latex]\\mathrm{sec}\\left(\\theta\\right) =-4,[\/latex] giving your answer in radians.<\/p>\n<\/div>\n<div id=\"fs-id1862382\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1862382\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1862382\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1625296\">We can begin with some algebra.<\/p>\n<div id=\"fs-id2713214\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}\\mathrm{sec}\\left(\\theta\\right) &=-4\\\\ \\frac{1}{\\mathrm{cos}\\left(\\theta\\right)}&=-4\\\\\\mathrm{cos}\\left(\\theta\\right) &=-\\frac{1}{4}\\end{align*}[\/latex]<\/div>\n<p id=\"fs-id2176861\">Check that the MODE is in radians. Now use the inverse cosine function.<\/p>\n<div id=\"fs-id1350281\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}\\mathrm{cos}^{-1}\\left(-\\frac{1}{4}\\right)&\\approx 1.8235\\\\ \\theta& \\approx 1.8235+2\\pi k\\\\ \\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1471997\">Since [latex]\\frac{\\pi }{2}\\approx 1.57[\/latex] and [latex]\\pi \\approx 3.14,[\/latex] we know that 1.8235 is between these two numbers, thus [latex]\\theta \\approx \\text{1}\\text{.8235}[\/latex] is in quadrant II. Cosine is also negative in quadrant III. Note that a calculator will only return an angle in quadrants I or II for the cosine function, since that is the range of the inverse cosine. See <a class=\"autogenerated-content\" href=\"#Figure_07_05_005\">Figure 1<\/a>.<\/p>\n<div id=\"Figure_07_05_005\" class=\"medium\">\n<div style=\"width: 380px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3896\/2019\/03\/07135354\/CNX_Precalc_Figure_07_05_005.jpg\" alt=\"Graph of angles theta =approx 1.8235, theta prime =approx pi - 1.8235 = approx 1.3181, and then theta prime = pi + 1.3181 = approx 4.4597\" width=\"370\" height=\"289\" \/><\/p>\n<p class=\"wp-caption-text\"><strong>Figure 1<\/strong><\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1631910\">So, we also need to find the measure of the angle in quadrant III. In quadrant III, the reference angle is [latex]\\theta^{\\prime} \\approx \\pi -\\text{1}\\text{.8235}\\approx \\text{1}\\text{.3181}\\text{.}[\/latex] The other solution in quadrant III is [latex]\\pi +\\text{1}\\text{.3181}\\approx \\text{4}\\text{.4597.}[\/latex]<\/p>\n<p id=\"fs-id1327393\">The solutions are [latex]1.8235\\pm2\\pi k[\/latex] and [latex]4.4597\\pm2\\pi k.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id2168685\" class=\"precalculus tryit\">\n<h3>Try it #3<\/h3>\n<div id=\"ti_07_05_03\">\n<div id=\"fs-id1868122\">\n<p id=\"fs-id2142937\">Solve [latex]\\mathrm{csc}\\left(\\theta\\right) =3.[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1562435\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1562435\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1562435\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1334489\">[latex]\\theta \\approx 0.33984\\pm2\\pi k[\/latex] and [latex]\\theta \\approx 2.80176\\pm2\\pi k[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id2461424\" class=\"bc-section section\">\n<h3>Solving Trigonometric Equations in Quadratic Form<\/h3>\n<p id=\"fs-id2590665\">Solving a <span class=\"no-emphasis\">quadratic equation<\/span> may be more complicated, but once again, we can use algebra as we would for any quadratic equation. Look at the pattern of the equation. Is there more than one trigonometric function in the equation, or is there only one? Which trigonometric function is squared? If there is only one function represented and one of the terms is squared, think about the standard form of a quadratic. Replace the trigonometric function with a variable such as [latex]x[\/latex] or [latex]u.[\/latex] If substitution makes the equation look like a quadratic equation, then we can use the same methods for solving quadratics to solve the trigonometric equations.<\/p>\n<div class=\"textbox examples\">\n<h3>How To<\/h3>\n<p id=\"fs-id1335125\"><strong>Given a trigonometric equation, solve using algebra<\/strong>.<\/p>\n<ol id=\"fs-id1753734\" type=\"1\">\n<li>Look for a pattern that suggests an algebraic property, such as the difference of squares or a factoring opportunity.<\/li>\n<li>Substitute the trigonometric expression with a single variable, such as [latex]x[\/latex] or [latex]u.[\/latex]<\/li>\n<li>Solve the equation the same way an algebraic equation would be solved.<\/li>\n<li>Substitute the trigonometric expression back in for the variable in the resulting expressions.<\/li>\n<li>Solve for the angle.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_07_05_04\" class=\"textbox examples\">\n<div id=\"fs-id1081089\">\n<div id=\"fs-id1290420\">\n<h3>Example 9:\u00a0 Solving a Trigonometric Equation in Quadratic Form Using the Square Root Property<\/h3>\n<p id=\"fs-id1354845\">Solve the problem exactly: [latex]2\\text{ }{\\mathrm{sin}}^{2}\\left(\\theta\\right) -1=0, \\text{ }0\\le \\theta <2\\pi .[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1552358\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1552358\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1552358\" class=\"hidden-answer\" style=\"display: none\">\nAs this problem is not easily factored, we will solve using the square root property. First, we use algebra to isolate [latex]\\mathrm{sin}\\left(\\theta\\right).[\/latex] Then we will find the angles.<\/p>\n<div id=\"fs-id1368172\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}2\\mathrm{sin}^{2}\\left(\\theta\\right) -1&=0\\\\ 2\\mathrm{sin}^{2}\\left(\\theta\\right)&=1&&\\text{Add 1 to both sides.} \\\\ \\mathrm{sin}^{2}\\left(\\theta\\right) &=\\frac{1}{2}&&\\text{Divide both sides by 2.}\\\\ \\sqrt{\\mathrm{sin}^{2}\\left(\\theta\\right)}&=\\pm\\sqrt{\\frac{1}{2}}&&\\text{Consider both positive and negative square root values.} \\\\ \\mathrm{sin}\\left(\\theta\\right) &=\\pm\\frac{1}{\\sqrt{2}}&&\\text{Recognize the value from a special right triangle.}\\\\ \\theta &=\\frac{\\pi }{4},\\text{ }\\frac{3\\pi }{4},\\text{ }\\frac{5\\pi }{4},\\text{ }\\frac{7\\pi }{4}&&\\text{Since we have + and -, we have answers in all 4 quadrants}\\end{align*}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_07_05_10\" class=\"textbox examples\">\n<div id=\"fs-id1335579\">\n<div id=\"fs-id1897958\">\n<h3>Example 10: Solving a Trigonometric Equation in Quadratic Form Using the Quadratic Equation<\/h3>\n<p id=\"fs-id2201479\">Solve the equation exactly: [latex]\\mathrm{cos}^{2}\\left(\\theta\\right)+3\\mathrm{cos}\\left(\\theta\\right)-1=0,\\text{ }0\\le \\theta <2\\pi .[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id761598\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id761598\">Show Solution<\/span><\/p>\n<div id=\"qfs-id761598\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1378461\">We begin by using substitution and replacing [latex]\\mathrm{cos}\\left(\\theta\\right)[\/latex] with [latex]u.[\/latex] It is not necessary to use substitution, but it may make the problem easier to solve visually. Let [latex]\\mathrm{cos}\\left(\\theta\\right) =u.[\/latex] We have<\/p>\n<div id=\"fs-id1875852\" class=\"unnumbered\" style=\"text-align: center\">[latex]{u}^{2}+3u-1=0[\/latex][latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id1356287\">The equation cannot be factored, so we will use the <span class=\"no-emphasis\">quadratic formula<\/span> [latex]u=\\frac{-b\\pm\\sqrt{{b}^{2}-4ac}}{2a},[\/latex] where [latex]a=1,\\text{ }b=3[\/latex] and [latex]c=-1.[\/latex]<\/p>\n<div id=\"fs-id2294578\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*} \\\\ u&=\\frac{-3\\pm\\sqrt{{\\left(3\\right)}^{2}-4\\left(1\\right)\\left(-1\\right)}}{2}\\\\ &=\\frac{-3\\pm\\sqrt{13}}{2}\\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\n<p>Replace [latex]u[\/latex] with [latex]\\mathrm{cos}\\left(\\theta\\right),[\/latex] and solve. Thus,<\/p>\n<div id=\"fs-id1408206\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*} \\mathrm{cos}\\left(\\theta\\right)&=\\frac{-3\\pm\\sqrt{13}}{2}\\\\\\theta&=\\mathrm{cos}^{-1}\\left(\\frac{-3+\\sqrt{13}}{2}\\right) \\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id1355539\">Note that only the + sign is used. This is because we get an error when we solve [latex]\\theta =\\mathrm{cos}^{-1}\\left(\\frac{-3-\\sqrt{13}}{2}\\right)[\/latex] on a calculator, since the domain of the inverse cosine function is [latex]\\left[-1,1\\right].[\/latex] Therefore the solution is<\/p>\n<div id=\"fs-id1236551\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}\\mathrm{cos}^{-1}\\left(\\frac{-3+\\sqrt{13}}{2}\\right)\\approx 1.26\\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id1298252\">This terminal side of the angle lies in quadrant I. Since cosine is also positive in quadrant IV, the second solution is<\/p>\n<div id=\"fs-id2160568\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}2\\pi -{\\mathrm{cos}}^{-1}\\left(\\frac{-3+\\sqrt{13}}{2}\\right)\\approx 5.02 \\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_07_05_11\" class=\"textbox examples\">\n<div id=\"fs-id1343020\">\n<div id=\"fs-id1334881\">\n<h3>Example 11:\u00a0 Solving a Trigonometric Equation in Quadratic Form by Factoring<\/h3>\n<p id=\"fs-id1403865\">Solve the equation exactly: [latex]2\\text{ }\\mathrm{sin}^{2}\\left(\\theta\\right) -5\\text{ }\\mathrm{sin}\\left(\\theta\\right)+3=0,\\text{ }0\\le \\theta \\le 2\\pi .[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id1218136\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1218136\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1218136\" class=\"hidden-answer\" style=\"display: none\">\nUsing grouping, this quadratic can be factored. Either make the real substitution, [latex]\\mathrm{sin}\\left(\\theta\\right) =u[\/latex] and factor [latex]2u^2-5u+3=0,[\/latex] or imagine it, as we factor:<\/p>\n<div id=\"fs-id2676123\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}\\textrm{Without the substitution }&& &&\\text{With the substitution}\\\\2\\text{ }\\mathrm{sin}^{2}\\left(\\theta\\right) -5\\text{ }\\mathrm{sin}\\left(\\theta\\right) +3&=0&& &2u^2-5u+3=0\\\\ \\left(2\\text{ }\\mathrm{sin}\\left(\\theta\\right)-3\\right)\\left(\\mathrm{sin}\\left(\\theta\\right) -1\\right)&=0 &&&\\left(2u-3\\right)\\left(u-1\\right)\\end{align*}[\/latex][latex]\\text{ }[\/latex][latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id1235701\">Now set each factor equal to zero and solve the two equations [latex]2\\text{ }\\mathrm{sin}\\left(\\theta\\right) -3=0[\/latex] and [latex]\\mathrm{sin}\\left(\\theta\\right) -1=0.[\/latex]<\/p>\n<p>For the first equation,<\/p>\n<div id=\"fs-id1634817\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}2\\text{ }\\mathrm{sin}\\left(\\theta\\right) -3&=0\\\\ 2\\text{ }\\mathrm{sin}\\left(\\theta\\right)& =3&&\\textrm{Add 3 to both sides.}\\\\ \\mathrm{sin}\\left(\\theta\\right)& =\\frac{3}{2}&&\\textrm{Divide both sides by 2.} \\\\ \\mathrm{sin}\\left(\\theta\\right)&\\ne \\frac{3}{2}&&\\frac{3}{2}\\textrm{ is not in the domain of the sine function.}\\end{align*}[\/latex][latex]\\text{ }[\/latex][latex]\\\\[\/latex]<\/div>\n<p>The first equation did not have any solution.\u00a0 To solve the second equation,<\/p>\n<div style=\"text-align: center\">[latex]\\begin{align*}\\mathrm{sin}\\left(\\theta\\right) -1&=0\\\\ \\mathrm{sin}\\left(\\theta\\right)&=1&&\\textrm{Add 1 to both sides.}\\\\\\theta&=\\frac{\\pi}{2}&&\\textrm{Sine equals one only for the quadrantal angle.}\\end{align*}[\/latex][latex]\\text{ }[\/latex][latex]\\\\[\/latex]<\/div>\n<div id=\"fs-id1218136\">\n<p>The only solution for this equation is [latex]\\theta=\\frac{\\pi}{2}.[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<div id=\"fs-id1220222\">\n<h3>Analysis<\/h3>\n<p>Make sure to check all solutions on the given domain as some factors have no solution. This is because\u00a0the range of the sine function is [latex]\\left[-1,1\\right].[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<div id=\"fs-id2123698\">\n<h3>Example 12: Solving an Equation Using an Identity<\/h3>\n<p id=\"fs-id2123703\">Solve the equation exactly using an identity: [latex]3\\text{ }\\mathrm{cos}\\left(\\theta\\right)+3=2\\text{ }{\\mathrm{sin}}^{2}\\left(\\theta\\right) ,\\text{ }0\\le \\theta <2\\pi .[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id2131508\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id2131508\">Show Solution<\/span><\/p>\n<div id=\"qfs-id2131508\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id2131510\">If we rewrite the right side, we can write the equation in terms of cosine.\u00a0 Recall that the Pythagorean Identity is [latex]\\mathrm{sin}^2\\left(\\theta\\right)+\\mathrm{cos}^2\\left(\\theta\\right)=1[\/latex] and can be solved for [latex]\\mathrm{sin}^{2}\\left(\\theta\\right)[\/latex] so [latex]\\mathrm{sin}^{2}\\left(\\theta\\right)=1-\\mathrm{cos}^{2}\\left(\\theta\\right).[\/latex]<\/p>\n<div id=\"fs-id3026861\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}3\\mathrm{cos}\\left(\\theta\\right)+3&=2\\mathrm{sin}^{2}\\left(\\theta\\right)\\\\ 3\\mathrm{cos}\\left(\\theta\\right)+3&=2\\left(1-\\mathrm{cos}^{2}\\left(\\theta\\right)\\right)&&\\text{Substitute for }\\mathrm{sin}^{2}\\left(\\theta\\right).\\\\\\end{align*}[\/latex]<\/div>\n<div><\/div>\n<div>Recognize that we now have a quadratic function in [latex]\\mathrm{cos}\\left(\\theta\\right).[\/latex]\u00a0 Factoring can be used to solve this quadratic equation.\u00a0 Below, the same steps used on the left are demonstrated on the right using the substitution method since the quadratic equation in that form may be easier to work with.\u00a0 For the substitution method, let [latex]u=\\mathrm{cos}\\left(\\theta\\right).[\/latex][latex]\\\\[\/latex]<\/div>\n<div><\/div>\n<div class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}\\textrm{Without the substitution }&& &\\text{With the substitution}\\\\3\\mathrm{cos}\\left(\\theta\\right)+3&=2-2\\mathrm{cos}^{2}\\left(\\theta\\right)&& 3u+3=2-2u^2\\\\ 2\\mathrm{cos}^{2}\\left(\\theta\\right)+3\\mathrm{cos}\\left(\\theta\\right)+1&=0&&2u^2+3u+1=0\\\\ \\left(2\\mathrm{cos}\\left(\\theta\\right)+1\\right)\\left(\\mathrm{cos}\\left(\\theta\\right)+1\\right)&=0&&\\left(2u+1\\right)\\left(u+1\\right)=0\\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\n<div><\/div>\n<div>Working with the factors using cosine, set each factor equal to zero and solve.[latex]\\\\[\/latex]<\/div>\n<div><\/div>\n<div style=\"text-align: center\"><span style=\"font-size: 1rem;text-align: initial\">[latex]\\begin{align*}2\\mathrm{cos}\\left(\\theta\\right)+1&=0 &\\mathrm{cos}\\left(\\theta\\right)+1&=0\\\\ \\mathrm{cos}\\left(\\theta\\right)&=-\\frac{1}{2}& \\mathrm{cos}\\left(\\theta\\right)&=-1\\\\\\theta&=\\frac{2\\pi}{3},\\text{ }\\frac{4\\pi}{3}\\text{ }\\text{ }\\text{ }\\text{ }\\text{ }&\\theta&=\\pi\\end{align*}[\/latex][latex]\\\\[\/latex]<\/span><\/div>\n<div class=\"unnumbered\" style=\"text-align: center\">\n<p>Again, remember that there are two quadrants where the [latex]\\mathrm{cos}\\left(\\theta\\right)=-\\frac{1}{2}.[\/latex]\u00a0 These are quadrants 2 and 3.\u00a0 We can find the acute reference angle of [latex]\\frac{\\pi}{3}[\/latex] and use that to generate the solutions in these quadrants.<\/p>\n<div id=\"fs-id2131508\">\n<p id=\"fs-id3483085\">Our solutions are [latex]\\frac{2\\pi}{3},\\text{ }\\frac{4\\pi }{3},\\text{ }\\pi.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1094265\" class=\"precalculus tryit\">\n<h3>Try it #4<\/h3>\n<div id=\"ti_07_05_04\">\n<div id=\"fs-id1213671\">\n<p id=\"fs-id1213672\">Solve [latex]\\mathrm{sin}^{2}\\left(\\theta\\right)=2\\mathrm{cos}\\left(\\theta\\right)+2,\\text{ }0\\le\\theta \\le2\\pi[\/latex] [Hint: Make a substitution to express the equation only in terms of cosine.]<\/p>\n<\/div>\n<div id=\"fs-id1333795\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1333795\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1333795\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id2262028\">[latex]\\mathrm{cos}\\left(\\theta\\right) =-1,\\text{ }\\theta =\\pi[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1892509\" class=\"precalculus tryit\">\n<h3>Try it #5<\/h3>\n<div id=\"ti_07_05_05\">\n<div id=\"fs-id2737020\">\n<p id=\"fs-id1743088\">Solve the quadratic equation [latex]2\\mathrm{cos}^{2}\\left(\\theta\\right) +\\mathrm{cos}\\left(\\theta\\right) =0.[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id3549256\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id3549256\">Show Solution<\/span><\/p>\n<div id=\"qfs-id3549256\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id3549257\">[latex]\\frac{\\pi }{2},\\text{ }\\frac{2\\pi }{3},\\text{ }\\frac{4\\pi }{3},\\text{ }\\frac{3\\pi }{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1597424\" class=\"bc-section section\">\n<h3>Solving Trigonometric Equations Using Fundamental Identities<\/h3>\n<p id=\"fs-id1396319\">While algebra can be used to solve a number of trigonometric equations, we can also use the fundamental identities because they make solving equations simpler. Remember that the techniques we use for solving are not the same as those for verifying identities. The basic rules of algebra apply here, as opposed to rewriting one side of the identity to match the other side. In the next example, we use two identities to simplify the equation.<\/p>\n<p>We will need to use some new identities in this section.\u00a0 They are called the\u00a0<strong>double-angle<\/strong> identities.\u00a0We will not take the time to show where the following identities come from. Keep in mind that there are other trigonometric identities that we have not covered in this material. If you are interested, you can look up the sum and difference formulas for sine and cosine, and use those to generate some other identities, including the ones shown below.<\/p>\n<div class=\"textbox definitions\">\n<h3>Definition<\/h3>\n<p id=\"fs-id1165137661589\">The double angle identities are:<\/p>\n<p style=\"text-align: center\">[latex]\\mathrm{sin}\\left(2\\theta\\right)=2\\mathrm{sin}\\left(\\theta\\right)\\mathrm{cos}\\left(\\theta\\right)[\/latex]<\/p>\n<p>[latex]\\\\[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align*}\\mathrm{cos}\\left(2\\theta\\right)&=\\mathrm{cos}^{2}\\left(\\theta\\right)-\\mathrm{sin}^{2}\\left(\\theta\\right)\\text{ or,}\\\\&=1-2\\mathrm{sin}^{2}\\left(\\theta\\right)\\text{ or,}\\\\&=2\\mathrm{cos}^{2}\\left(\\theta\\right)-1\\end{align*}[\/latex]<\/p>\n<\/div>\n<div id=\"Example_07_05_15\" class=\"textbox examples\">\n<div id=\"fs-id1916902\">\n<div id=\"fs-id1916904\">\n<h3>Example 13: Solving the Equation Using a Double-Angle Formula<\/h3>\n<p id=\"fs-id1916909\">Solve the equation exactly using a double-angle formula: [latex]\\mathrm{cos}\\left(2\\theta\\right)=\\mathrm{cos}\\left(\\theta\\right).[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id697866\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id697866\">Show Solution<\/span><\/p>\n<div id=\"qfs-id697866\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id697868\">We have three choices of expressions to substitute for the double-angle of cosine. As it is simpler to solve for one trigonometric function at a time, we will choose the double-angle identity involving only cosine:<\/p>\n<div id=\"fs-id2070377\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}\\mathrm{cos}\\left(2\\theta\\right)&=\\mathrm{cos}\\left(\\theta\\right)\\\\ 2\\mathrm{cos}^{2}\\left(\\theta\\right)-1&=\\mathrm{cos}\\left(\\theta\\right)&&\\text{Replace left hand side with identity.}\\\\ 2\\mathrm{cos}^{2}\\left(\\theta\\right)-\\mathrm{cos}\\left(\\theta\\right)-1&=0&&\\text{Move all terms to one side.}\\\\ \\left(2\\mathrm{cos}\\left(\\theta\\right)+1\\right)\\left(\\mathrm{cos}\\left(\\theta\\right)-1\\right)&=0&&\\text{Factor the left hand side.}\\end{align*}[\/latex]<\/div>\n<p>[latex]\\\\[\/latex]<\/p>\n<p>Set the factors equal to zero and solve.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align*}2\\mathrm{cos}\\left(\\theta\\right)+1&=0&\\mathrm{cos}\\left(\\theta\\right)-1&=0\\\\\\mathrm{cos}\\left(\\theta\\right)&=-\\frac{1}{2}&\\mathrm{cos}\\left(\\theta\\right)&=1\\\\\\theta&=\\frac{2\\pi}{3}\\pm2\\pi k&\\theta&=0\\pm2\\pi k\\\\\\theta&=\\frac{4\\pi}{3}\\pm2\\pi k\\end{align*}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1518803\" class=\"bc-section section\">\n<h3>Solving Trigonometric Equations with Multiple Angles<\/h3>\n<p id=\"fs-id1518808\">Sometimes it is not possible to solve a trigonometric equation with identities that have a multiple angle, such as [latex]\\mathrm{sin}\\left(2x\\right)[\/latex] or [latex]\\mathrm{cos}\\left(3x\\right).[\/latex] When confronted with these equations, recall that [latex]y=\\mathrm{sin}\\left(2x\\right)[\/latex] is a <span class=\"no-emphasis\">horizontal compression<\/span> by a factor of 2 of the function [latex]y=\\mathrm{sin}\\left(x\\right).[\/latex]\u00a0 On an interval of [latex]2\\pi,[\/latex] we can graph two periods of [latex]y=\\mathrm{sin}\\left(2x\\right),[\/latex] as opposed to one cycle of [latex]y=\\mathrm{sin}\\left(x\\right).[\/latex] This compression of the graph leads us to believe there may be twice as many <em>x<\/em>-intercepts or solutions to [latex]\\mathrm{sin}\\left(2x\\right)=0[\/latex] compared to [latex]\\mathrm{sin}\\left(x\\right)=0.[\/latex] This information will help us solve the similar type of equation shown in the example.<\/p>\n<div id=\"Example_07_05_17\" class=\"textbox examples\">\n<div id=\"fs-id2223187\">\n<div id=\"fs-id2223189\">\n<h3>Example 14: Solving a Multiple Angle Trigonometric Equation<\/h3>\n<p id=\"fs-id2223194\">Solve exactly: [latex]\\mathrm{cos}\\left(2x\\right)=\\frac{1}{2}[\/latex] on [latex]\\left[0,2\\pi \\right).[\/latex]<\/p>\n<\/div>\n<div id=\"fs-id2740923\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id2740923\">Show Solution<\/span><\/p>\n<div id=\"qfs-id2740923\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id2740926\">We can see that this equation is the standard equation with a multiple of an angle. If [latex]\\mathrm{cos}\\left(\\theta\\right)=\\frac{1}{2},[\/latex] we know [latex]\\theta[\/latex] is in quadrants I and IV. While [latex]\\theta =\\mathrm{cos}^{-1}\\left(\\frac{1}{2}\\right)[\/latex] will only yield solutions in quadrants I and II because of the range of the inverse cosine function, we recognize that the solutions to the equation [latex]\\mathrm{cos}\\left(\\theta\\right)=\\frac{1}{2}[\/latex] will be in quadrants I and IV using ideas from our unit circle.<\/p>\n<p id=\"fs-id3162261\">Therefore, the possible angles are [latex]\\theta=\\frac{\\pi }{3}[\/latex] and [latex]\\theta=\\frac{5\\pi}{3}.[\/latex] So, [latex]2x=\\frac{\\pi}{3}[\/latex] or [latex]2x=\\frac{5\\pi}{3},[\/latex] which means that [latex]x=\\frac{\\pi}{6}[\/latex] or [latex]x=\\frac{5\\pi}{6}.[\/latex] [latex]\\\\[\/latex]Does this make sense? Yes, because [latex]\\mathrm{cos}\\left(2\\left(\\frac{\\pi}{6}\\right)\\right)=\\mathrm{cos}\\left(\\frac{\\pi}{3}\\right)=\\frac{1}{2}.[\/latex]<\/p>\n<p id=\"fs-id1966562\">Are there any other possible answers? Let us return to our first step.<\/p>\n<p id=\"fs-id1966566\">In quadrant I, [latex]2x=\\frac{\\pi}{3},[\/latex] so [latex]x=\\frac{\\pi}{6}[\/latex] as noted. Let us revolve around the circle again:<\/p>\n<div id=\"fs-id1893406\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}2x&=\\frac{\\pi}{3}+2\\pi \\\\ 2x&=\\frac{\\pi }{3}+\\frac{6\\pi}{3}\\\\ 2x&=\\frac{7\\pi }{3}\\\\x&=\\frac{7\\pi}{6}\\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id1805664\">One more rotation yields<\/p>\n<div id=\"fs-id1805667\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}2x&=\\frac{\\pi }{3}+4\\pi \\\\2x&=\\frac{\\pi}{3}+\\frac{12\\pi}{3}\\\\ 2x&=\\frac{13\\pi}{3}\\\\x&=\\frac{13\\pi}{6}\\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id3654355\">[latex]x=\\frac{13\\pi}{6}>2\\pi,[\/latex] so this value for [latex]x[\/latex] is larger than [latex]2\\pi,[\/latex] so it is not a solution on [latex]\\left[0,2\\pi \\right).[\/latex]<\/p>\n<p id=\"fs-id2748825\">In quadrant IV, [latex]2x=\\frac{5\\pi }{3},[\/latex] so [latex]x=\\frac{5\\pi }{6}[\/latex] as noted. Let us revolve around the circle again:<\/p>\n<div id=\"fs-id1154337\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}2x&=\\frac{5\\pi}{3}+2\\pi \\\\2x&=\\frac{5\\pi }{3}+\\frac{6\\pi }{3} \\\\ 2x&=\\frac{11\\pi }{3}\\\\x&=\\frac{11\\pi}{6} \\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id2123818\">One more rotation yields<\/p>\n<div id=\"fs-id2123822\" class=\"unnumbered\" style=\"text-align: center\">[latex]\\begin{align*}2x&=\\frac{5\\pi }{3}+4\\pi \\\\ 2x&=\\frac{5\\pi }{3}+\\frac{12\\pi }{3} \\\\ 2x&=\\frac{17\\pi }{3}\\\\x&=\\frac{17\\pi}{6}\\end{align*}[\/latex][latex]\\\\[\/latex]<\/div>\n<p id=\"fs-id1217412\">[latex]x=\\frac{17\\pi }{6}>2\\pi ,[\/latex] so this value for [latex]x[\/latex] is larger than [latex]2\\pi,[\/latex] which means it is not a solution on [latex]\\left[0,2\\pi \\right).[\/latex]<\/p>\n<p id=\"fs-id1676669\">Our solutions are [latex]\\frac{\\pi}{6},\\text{ }\\frac{5\\pi }{6},\\text{ }\\frac{7\\pi }{6},[\/latex] and [latex]\\frac{11\\pi }{6}.[\/latex] Note that whenever we solve a problem in the form of [latex]\\mathrm{sin}\\left(nx\\right)=c[\/latex] or [latex]\\mathrm{cos}\\left(nx\\right)=c,[\/latex] we must go around the unit circle [latex]{n}[\/latex] times.<\/p>\n<p>We can see this easily if we first write the solution for [latex]2x[\/latex] in the general form.\u00a0 One of these equations is shown below:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align*}2x&=\\frac{\\pi}{3}+2\\pi\\cdot k\\\\x&=\\frac{\\pi}{6}+\\pi \\cdot k&&\\text{Divide both sides by 2.}\\end{align*}[\/latex][latex]\\\\[\/latex]<\/p>\n<p>We can now see that adding [latex]\\pi[\/latex] to [latex]\\frac{\\pi}{6}[\/latex] gives us\u00a0[latex]\\frac{7\\pi}{6}[\/latex] and that adding [latex]2\\cdot\\pi[\/latex] would give us\u00a0[latex]\\frac{13\\pi}{6}[\/latex]\u00a0 which is larger than [latex]2\\pi.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id2267036\" class=\"precalculus media\">\n<p id=\"fs-id2267042\">Access these online resources for additional instruction and practice with solving trigonometric equations.<\/p>\n<ul id=\"fs-id2267047\">\n<li><a href=\"http:\/\/openstax.org\/l\/solvetrigeqI\">Solving Trigonometric Equations I<\/a><\/li>\n<li><a href=\"https:\/\/www.youtube.com\/watch?v=ABKO3ta_Azw\">Solving Trigonometric Equations II<\/a><\/li>\n<li><a href=\"http:\/\/openstax.org\/l\/solvetrigeqIII\">Solving Trigonometric Equations III<\/a><\/li>\n<li><a href=\"http:\/\/openstax.org\/l\/solvetrigeqIV\">Solving Trigonometric Equations IV<\/a><\/li>\n<li><a href=\"http:\/\/openstax.org\/l\/solvetrigeqV\">Solving Trigonometric Equations V<\/a><\/li>\n<li><a href=\"http:\/\/openstax.org\/l\/solvetrigeqVI\">Solving Trigonometric Equations VI<\/a><\/li>\n<\/ul>\n<\/div>\n<\/div>\n<div id=\"fs-id2267085\" class=\"textbox key-takeaways\">\n<h3>Key Concepts<\/h3>\n<ul id=\"fs-id2267091\">\n<li>When solving linear trigonometric equations, we can use algebraic techniques just as we do solving algebraic equations. Look for patterns, like the difference of squares, quadratic form, or an expression that lends itself well to substitution.<\/li>\n<li>Equations involving a single trigonometric function can be solved or verified using the unit circle.<\/li>\n<li>We can also solve trigonometric equations using a graphing calculator.<\/li>\n<li>Many equations appear quadratic in form. We can use substitution to make the equation appear simpler, and then use the same techniques we use solving an algebraic quadratic: factoring, the quadratic formula, etc.<\/li>\n<li>We can also use the identities to solve trigonometric equation.<\/li>\n<li>We can use substitution to solve a multiple-angle trigonometric equation, which is a compression of a standard trigonometric function. We will need to take the compression into account and verify that we have found all solutions on the given interval.<\/li>\n<\/ul>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-790\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Solving Trigonometric Equations. <strong>Authored by<\/strong>: Douglas Hoffman. <strong>Provided by<\/strong>: Openstax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/cnx.org\/contents\/8si1Yf2B@2.21:aeVxcRIM@7\/Solving-Trigonometric-Equations\">https:\/\/cnx.org\/contents\/8si1Yf2B@2.21:aeVxcRIM@7\/Solving-Trigonometric-Equations<\/a>. <strong>Project<\/strong>: Essential Precalcus, Part 2. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":311,"menu_order":16,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Solving Trigonometric Equations\",\"author\":\"Douglas Hoffman\",\"organization\":\"Openstax\",\"url\":\"https:\/\/cnx.org\/contents\/8si1Yf2B@2.21:aeVxcRIM@7\/Solving-Trigonometric-Equations\",\"project\":\"Essential Precalcus, Part 2\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-790","chapter","type-chapter","status-publish","hentry"],"part":478,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/chapters\/790","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/wp\/v2\/users\/311"}],"version-history":[{"count":40,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/chapters\/790\/revisions"}],"predecessor-version":[{"id":3246,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/chapters\/790\/revisions\/3246"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/parts\/478"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/chapters\/790\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/wp\/v2\/media?parent=790"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=790"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/wp\/v2\/contributor?post=790"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-precalculus\/wp-json\/wp\/v2\/license?post=790"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}