{"id":196,"date":"2021-06-04T18:09:58","date_gmt":"2021-06-04T18:09:58","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/gsu-precalculus\/chapter\/solving-trigonometric-equations\/"},"modified":"2021-06-14T23:50:58","modified_gmt":"2021-06-14T23:50:58","slug":"solving-trigonometric-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/gsu-precalculus\/chapter\/solving-trigonometric-equations\/","title":{"raw":"4.4 Solving Trigonometric Equations","rendered":"4.4 Solving Trigonometric Equations"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li style=\"font-weight: 400;\">Solve equations involving a single trigonometric function.<\/li>\r\n \t<li style=\"font-weight: 400;\">Solve trigonometric equations using a calculator.<\/li>\r\n \t<li style=\"font-weight: 400;\">Solve trigonometric equations that involve factoring.<\/li>\r\n \t<li style=\"font-weight: 400;\">Solve trigonometric equations using fundamental identities.<\/li>\r\n \t<li style=\"font-weight: 400;\">Solve trigonometric equations with multiple angles.<\/li>\r\n \t<li style=\"font-weight: 400;\">Solve right triangle problems.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Solving Linear Trigonometric Equations in Sine and Cosine<\/h2>\r\nTrigonometric 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. 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. Additionally, like rational equations, the domain of the function must be considered before we assume that any solution is valid. The <strong>period<\/strong> 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. 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. Recall the rule that gives the format for stating all possible solutions for a function where the period is [latex]2\\pi :[\/latex]\r\n<div style=\"text-align: center;\">[latex]\\sin \\theta =\\sin \\left(\\theta \\pm 2k\\pi \\right)[\/latex]<\/div>\r\nThere 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.\r\n<div class=\"textbox shaded\">\r\n<h3>Example 1: Solving a Linear Trigonometric Equation Involving the Cosine Function<\/h3>\r\nFind all possible exact solutions for the equation [latex]\\cos \\theta =\\frac{1}{2}[\/latex].\r\n\r\n[reveal-answer q=\"84784\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"84784\"]\r\n\r\nFrom the <strong>unit circle<\/strong>, we know that\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\cos \\theta =\\frac{1}{2} \\\\ \\theta =\\frac{\\pi }{3},\\frac{5\\pi }{3} \\end{gathered}[\/latex]<\/p>\r\nThese are the solutions in the interval [latex]\\left[0,2\\pi \\right][\/latex]. All possible solutions are given by\r\n<p style=\"text-align: center;\">[latex]\\theta =\\frac{\\pi }{3}\\pm 2k\\pi \\text{ and }\\theta =\\frac{5\\pi }{3}\\pm 2k\\pi [\/latex]<\/p>\r\nwhere [latex]k[\/latex] is an integer.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 2: Solving a Linear Equation Involving the Sine Function<\/h3>\r\nFind all possible exact solutions for the equation [latex]\\sin t=\\frac{1}{2}[\/latex].\r\n\r\n[reveal-answer q=\"535703\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"535703\"]\r\n\r\nSolving for all possible values of <em>t<\/em> means that solutions include angles beyond the period of [latex]2\\pi [\/latex]. From the unit circle, we can see that the solutions are [latex]t=\\frac{\\pi }{6}[\/latex] and [latex]t=\\frac{5\\pi }{6}[\/latex]. But the problem is asking for all possible values that solve the equation. Therefore, the answer is\r\n<p style=\"text-align: center;\">[latex]t=\\frac{\\pi }{6}\\pm 2\\pi k\\text{ and }t=\\frac{5\\pi }{6}\\pm 2\\pi k[\/latex]<\/p>\r\nwhere [latex]k[\/latex] is an integer.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given a trigonometric equation, solve using algebra.<\/h3>\r\n<ul>\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<\/ul>\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 3: Solve the Trigonometric Equation in Linear Form<\/h3>\r\nSolve the equation exactly: [latex]2\\cos \\theta -3=-5,0\\le \\theta &lt;2\\pi [\/latex].\r\n\r\n[reveal-answer q=\"939405\"]Show Solution[\/reveal-answer]\r\n<p style=\"text-align: left;\">[hidden-answer a=\"939405\"]<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}2\\cos \\theta -3=-5 \\\\ \\cos \\theta =-2 \\\\ \\cos \\theta =-1 \\\\ \\theta =\\pi \\end{gathered}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nSolve exactly the following linear equation on the interval [latex]\\left[0,2\\pi \\right):2\\sin x+1=0[\/latex].\r\n\r\n[reveal-answer q=\"500341\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"500341\"]\r\n\r\n[latex]x=\\frac{7\\pi }{6},\\frac{11\\pi }{6}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]149871[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Solve Trigonometric Equations Using a Calculator<\/h2>\r\nNot 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.\r\n<div class=\"textbox shaded\">\r\n<h3>Example 4: Using a Calculator to Solve a Trigonometric Equation Involving Sine<\/h3>\r\nUse a calculator to solve the equation [latex]\\sin \\theta =0.8[\/latex], where [latex]\\theta [\/latex] is in radians.\r\n\r\n[reveal-answer q=\"522194\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"522194\"]\r\n<h3><\/h3>\r\nMake 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]{\\sin }^{-1}[\/latex] function. What is shown on the screen is [latex]{\\sin}^{-1}[\/latex](. The calculator is ready for the input within the parentheses. For this problem, we enter [latex]{\\sin }^{-1}\\left(0.8\\right)[\/latex], and press ENTER. Thus, to four decimals places,\r\n<p style=\"text-align: center;\">[latex]{\\sin }^{-1}\\left(0.8\\right)\\approx 0.9273[\/latex]<\/p>\r\nThis is the solution in quadrant I. There is also a solution in quadrant II. To find this we subtract [latex]\/pi - 0.9273 \\approx 2.2143 [\/latex]\r\n\r\nThe general solution is\r\n<p style=\"text-align: center;\">[latex]\\theta \\approx 0.9273\\pm 2\\pi k \\text{ and } \\theta \\approx 2.2143 \\pm 2\\pi k[\/latex]<\/p>\r\nThe angle measurement in degrees is\r\n<p style=\"text-align: center;\">[latex]\\begin{align} \\theta &amp;\\approx {53.1}^{\\circ } \\\\ \\theta &amp;\\approx {180}^{\\circ }-{53.1}^{\\circ } \\\\ &amp;\\approx {126.9}^{\\circ } \\end{align}[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\nNote that a calculator will only return an angle in quadrants I or IV for the sine function, since that is the range of the inverse sine. The other angle is obtained by using [latex]\\pi -\\theta [\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 5: Using a Calculator to Solve a Trigonometric Equation Involving Secant<\/h3>\r\nUse a calculator to solve the equation [latex]\\sec \\theta =-4[\/latex], giving your answer in radians.\r\n\r\n[reveal-answer q=\"209133\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"209133\"]\r\n\r\nWe can begin with some algebra.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\sec \\theta =-4\\\\ \\frac{1}{\\cos \\theta }=-4\\\\ \\cos \\theta =-\\frac{1}{4}\\end{gathered}[\/latex]<\/p>\r\nCheck that the MODE is in radians. Now use the inverse cosine function.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}{\\cos }^{-1}\\left(-\\frac{1}{4}\\right)\\approx 1.8235 \\\\ \\theta \\approx 1.8235+2\\pi k \\end{gathered}[\/latex]<\/p>\r\nSince [latex]\\frac{\\pi }{2}\\approx 1.57[\/latex] and [latex]\\pi \\approx 3.14[\/latex], 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.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164111\/CNX_Precalc_Figure_07_05_0052.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=\"487\" height=\"380\" \/> <b>Figure 2.<\/b>[\/caption]\r\n\r\nSo, we also need to find the measure of the angle in quadrant III. In quadrant III, the reference angle is [latex]\\theta \\text{ }\\text{ }\\text{'}\\approx \\pi -\\text{1}\\text{.8235}\\approx \\text{1}\\text{.3181}\\text{.}[\/latex] The other solution in quadrant III is [latex]\\theta \\text{ }\\text{ }\\text{'}\\approx \\pi +\\text{1}\\text{.3181}\\approx \\text{4}\\text{.4597}\\text{.}[\/latex]\r\n\r\nThe solutions are [latex]\\theta \\approx 1.8235\\pm 2\\pi k[\/latex] and [latex]\\theta \\approx 4.4597\\pm 2\\pi k[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nSolve [latex]\\cos \\theta =-0.2[\/latex].\r\n\r\n[reveal-answer q=\"145806\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"145806\"]\r\n\r\n[latex]\\theta \\approx 1.7722\\pm 2\\pi k[\/latex] and [latex]\\theta \\approx 4.5110\\pm 2\\pi k[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]149873[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Solving Equations Involving a Single Trigonometric Function<\/h2>\r\nWhen we are given equations that involve only one of the six trigonometric functions, their solutions involve using algebraic techniques and 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 reciprocal function, and solve for the angles using the function. 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.\r\n<div class=\"textbox shaded\">\r\n<h3>Example 6: Solving a Problem Involving a Single Trigonometric Function<\/h3>\r\nSolve the problem exactly: [latex]2{\\sin }^{2}\\theta -1=0,0\\le \\theta &lt;2\\pi [\/latex].\r\n\r\n[reveal-answer q=\"467313\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"467313\"]\r\n\r\nAs this problem is not easily factored, we will solve using the square root property. First, we use algebra to isolate [latex]\\sin \\theta [\/latex]. Then we will find the angles.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}2{\\sin }^{2}\\theta -1=0 \\\\ 2{\\sin }^{2}\\theta =1 \\\\ {\\sin }^{2}\\theta =\\frac{1}{2} \\\\ \\sqrt{{\\sin }^{2}\\theta }=\\pm \\sqrt{\\frac{1}{2}} \\\\ \\sin \\theta =\\pm \\frac{1}{\\sqrt{2}}=\\pm \\frac{\\sqrt{2}}{2} \\\\ \\theta =\\frac{\\pi }{4},\\frac{3\\pi }{4},\\frac{5\\pi }{4},\\frac{7\\pi }{4} \\end{gathered}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 7: Solving a Trigonometric Equation Involving Cosecant<\/h3>\r\nSolve the following equation exactly: [latex]\\csc \\theta =-2,0\\le \\theta &lt;4\\pi [\/latex].\r\n\r\n[reveal-answer q=\"605306\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"605306\"]\r\n\r\nWe want all values of [latex]\\theta [\/latex] for which [latex]\\csc \\theta =-2[\/latex] over the interval [latex]0\\le \\theta &lt;4\\pi [\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\csc \\theta =-2 \\\\ \\frac{1}{\\sin \\theta }=-2 \\\\ \\sin \\theta =-\\frac{1}{2} \\\\ \\theta =\\frac{7\\pi }{6},\\frac{11\\pi }{6},\\frac{19\\pi }{6},\\frac{23\\pi }{6} \\end{gathered}[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\nAs [latex]\\sin \\theta =-\\frac{1}{2}[\/latex], notice that all four solutions are in the third and fourth quadrants.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 8: Solving an Equation Involving Tangent<\/h3>\r\nSolve the equation exactly: [latex]\\tan \\left(\\theta -\\frac{\\pi }{2}\\right)=1,0\\le \\theta &lt;2\\pi [\/latex].\r\n\r\n[reveal-answer q=\"484899\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"484899\"]\r\n\r\nRecall 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]\\tan \\left(\\frac{\\pi }{4}\\right)=1[\/latex], then\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\theta -\\frac{\\pi }{2}=\\frac{\\pi }{4}\\\\ \\theta =\\frac{3\\pi }{4}\\pm k\\pi \\end{gathered}[\/latex]<\/p>\r\nOver the interval [latex]\\left[0,2\\pi \\right)[\/latex], we have two solutions:\r\n<p style=\"text-align: center;\">[latex]\\theta =\\frac{3\\pi }{4}\\text{ and }\\theta =\\frac{3\\pi }{4}+\\pi =\\frac{7\\pi }{4}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nFind all solutions for [latex]\\tan x=\\sqrt{3}[\/latex].\r\n\r\n[reveal-answer q=\"629684\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"629684\"]\r\n\r\n[latex]\\frac{\\pi }{3}\\pm \\pi k[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]173739[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 9: Identify all Solutions to the Equation Involving Tangent<\/h3>\r\nIdentify all exact solutions to the equation [latex]2\\left(\\tan x+3\\right)=5+\\tan x,0\\le x&lt;2\\pi [\/latex].\r\n\r\n[reveal-answer q=\"949694\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"949694\"]\r\n\r\nWe can solve this equation using only algebra. Isolate the expression [latex]\\tan x[\/latex] on the left side of the equals sign.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered} 2\\left(\\tan x\\right)+2\\left(3\\right) =5+\\tan x \\\\ 2\\tan x+6 =5+\\tan x \\\\ 2\\tan x-\\tan x =5 - 6 \\\\ \\tan x =-1\\end{gathered}[\/latex]<\/p>\r\nThere 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].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Solving Trigonometric Equations in Quadratic Form<\/h2>\r\nSolving a <strong>quadratic equation<\/strong> 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.\r\n<div class=\"textbox shaded\">\r\n<h3>Example 10: Solving a Trigonometric Equation in Quadratic Form<\/h3>\r\nSolve the equation exactly: [latex]{\\cos }^{2}\\theta +3\\cos \\theta -1=0,0\\le \\theta &lt;2\\pi[\/latex].\r\n\r\n[reveal-answer q=\"563509\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"563509\"]\r\n\r\nWe begin by using substitution and replacing cos [latex]\\theta[\/latex] with [latex]x[\/latex]. It is not necessary to use substitution, but it may make the problem easier to solve visually. Let [latex]\\cos \\theta =x[\/latex]. We have\r\n<p style=\"text-align: center;\">[latex]{x}^{2}+3x - 1=0[\/latex]<\/p>\r\nThe equation cannot be factored, so we will use the <strong>quadratic formula<\/strong> [latex]x=\\frac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a}[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align} x&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]<\/p>\r\nReplace [latex]x[\/latex] with [latex]\\cos \\theta[\/latex], and solve. Thus,\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered} \\cos \\theta =\\frac{-3\\pm \\sqrt{13}}{2}\\theta ={\\cos }^{-1}\\left(\\frac{-3+\\sqrt{13}}{2}\\right)\\end{gathered}[\/latex]<\/p>\r\nNote that only the + sign is used. This is because we get an error when we solve [latex]\\theta ={\\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]. However, there is a second solution:\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\theta &amp;={\\cos }^{-1}\\left(\\frac{-3+\\sqrt{13}}{2}\\right) \\\\ &amp;\\approx 1.26 \\end{align}[\/latex]<\/p>\r\nThis terminal side of the angle lies in quadrant I. Since cosine is also positive in quadrant IV, the second solution is\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\theta &amp;=2\\pi -{\\cos }^{-1}\\left(\\frac{-3+\\sqrt{13}}{2}\\right) \\\\ &amp;\\approx 5.02 \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 11: Solving a Trigonometric Equation in Quadratic Form by Factoring<\/h3>\r\nSolve the equation exactly: [latex]2{\\sin }^{2}\\theta -5\\sin \\theta +3=0,0\\le \\theta \\le 2\\pi[\/latex].\r\n\r\n[reveal-answer q=\"693370\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"693370\"]\r\n\r\nUsing grouping, this quadratic can be factored. Either make the real substitution, [latex]\\sin \\theta =u[\/latex], or imagine it, as we factor:\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}2{\\sin }^{2}\\theta -5\\sin \\theta +3=0 \\\\ \\left(2\\sin \\theta -3\\right)\\left(\\sin \\theta -1\\right)=0 \\end{gathered}[\/latex]<\/p>\r\nNow set each factor equal to zero.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}2\\sin \\theta -3=0 \\\\ 2\\sin \\theta =3 \\\\ \\sin \\theta =\\frac{3}{2} \\\\ \\text{ } \\\\ \\sin \\theta -1=0 \\\\ \\sin \\theta =1 \\end{gathered}[\/latex]<\/p>\r\nNext solve for [latex]\\theta :\\sin \\theta \\ne \\frac{3}{2}[\/latex], as the range of the sine function is [latex]\\left[-1,1\\right][\/latex]. However, [latex]\\sin \\theta =1[\/latex], giving the solution [latex]\\theta =\\frac{\\pi }{2}[\/latex].\r\n<h4>Analysis of the Solution<\/h4>\r\nMake sure to check all solutions on the given domain as some factors have no solution.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nSolve [latex]{\\sin }^{2}\\theta =2\\cos \\theta +2,0\\le \\theta \\le 2\\pi[\/latex]. [Hint: Make a substitution to express the equation only in terms of cosine.]\r\n\r\n[reveal-answer q=\"608693\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"608693\"]\r\n\r\n[latex]\\cos \\theta =-1,\\theta =\\pi [\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]149010[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 12: Solving a Trigonometric Equation Using Algebra<\/h3>\r\nSolve exactly:\r\n<p style=\"text-align: center;\">[latex]2{\\sin }^{2}\\theta +\\sin \\theta =0;0\\le \\theta &lt;2\\pi[\/latex]<\/p>\r\n[reveal-answer q=\"204542\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"204542\"]\r\n\r\nThis problem should appear familiar as it is similar to a quadratic. Let [latex]\\sin \\theta =x[\/latex]. The equation becomes [latex]2{x}^{2}+x=0[\/latex]. We begin by factoring:\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}2{x}^{2}+x=0\\\\ x\\left(2x+1\\right)=0\\end{gathered}[\/latex]<\/p>\r\nSet each factor equal to zero.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}x=0 \\\\ 2x+1=0 \\\\ x=-\\frac{1}{2} \\end{gathered}[\/latex]<\/p>\r\nThen, substitute back into the equation the original expression [latex]\\sin \\theta[\/latex] for [latex]x[\/latex]. Thus,\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\sin \\theta =0 \\\\ \\theta =0,\\pi \\\\ \\text{ } \\\\ \\sin \\theta =-\\frac{1}{2} \\\\ \\theta =\\frac{7\\pi }{6},\\frac{11\\pi }{6} \\end{gathered}[\/latex]<\/p>\r\nThe solutions within the domain [latex]0\\le \\theta &lt;2\\pi[\/latex] are [latex]\\theta =0,\\pi ,\\frac{7\\pi }{6},\\frac{11\\pi }{6}[\/latex].\r\n\r\nIf we prefer not to substitute, we can solve the equation by following the same pattern of factoring and setting each factor equal to zero.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}2{\\sin }^{2}\\theta +\\sin \\theta =0 \\\\ \\sin \\theta \\left(2\\sin \\theta +1\\right)=0 \\\\ \\sin \\theta =0 \\\\ \\theta =0,\\pi \\\\ \\text{ } \\\\ 2\\sin \\theta +1=0 \\\\ 2\\sin \\theta =-1 \\\\ \\sin \\theta =-\\frac{1}{2} \\\\ \\theta =\\frac{7\\pi }{6},\\frac{11\\pi }{6} \\end{gathered}[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\nWe can see the solutions on the graph in Figure 3. On the interval [latex]0\\le \\theta &lt;2\\pi[\/latex], the graph crosses the <em>x-<\/em>axis four times, at the solutions noted. Notice that trigonometric equations that are in quadratic form can yield up to four solutions instead of the expected two that are found with quadratic equations. In this example, each solution (angle) corresponding to a positive sine value will yield two angles that would result in that value.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164113\/CNX_Precalc_Figure_07_05_0042.jpg\" alt=\"Graph of 2*(sin(theta))^2 + sin(theta) from 0 to 2pi. Zeros are at 0, pi, 7pi\/6, and 11pi\/6.\" width=\"731\" height=\"256\" \/> <b>Figure 3<\/b>[\/caption]\r\n\r\nWe can verify the solutions on the <a href=\"https:\/\/courses.lumenlearning.com\/precalctwoxmaster\/chapter\/use-sum-and-difference-formulas-for-cosine\/\" target=\"_blank\" rel=\"noopener\">unit circle<\/a> in Sum and Difference Identities\u00a0as well.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 13: Solving a Trigonometric Equation Quadratic in Form<\/h3>\r\nSolve the equation quadratic in form exactly: [latex]2{\\sin }^{2}\\theta -3\\sin \\theta +1=0,0\\le \\theta &lt;2\\pi[\/latex].\r\n\r\n[reveal-answer q=\"699307\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"699307\"]\r\n\r\nWe can factor using grouping. Solution values of [latex]\\theta[\/latex] can be found on the unit circle:\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\left(2\\sin \\theta -1\\right)\\left(\\sin \\theta -1\\right)=0 \\\\ 2\\sin \\theta -1=0 \\\\ \\sin \\theta =\\frac{1}{2} \\\\ \\theta =\\frac{\\pi }{6},\\frac{5\\pi }{6} \\\\ \\text{ } \\\\ \\sin \\theta =1 \\\\ \\theta =\\frac{\\pi }{2} \\end{gathered}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\nSolve the quadratic equation [latex]2{\\cos }^{2}\\theta +\\cos \\theta =0[\/latex].\r\n\r\n[reveal-answer q=\"736189\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"736189\"]\r\n\r\n[latex]\\frac{\\pi }{2},\\frac{2\\pi }{3},\\frac{4\\pi }{3},\\frac{3\\pi }{2}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Solving Trigonometric Equations Using Fundamental Identities<\/h2>\r\nWhile 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.\r\n<div class=\"textbox shaded\">\r\n<h3>Example 14: Use Identities to Solve an Equation<\/h3>\r\nUse identities to solve exactly the trigonometric equation over the interval [latex]0\\le x&lt;2\\pi [\/latex].\r\n<p style=\"text-align: center;\">[latex]\\cos x\\cos \\left(2x\\right)+\\sin x\\sin \\left(2x\\right)=\\frac{\\sqrt{3}}{2}[\/latex]<\/p>\r\n[reveal-answer q=\"607762\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"607762\"]\r\n\r\nNotice that the left side of the equation is the difference formula for cosine.\r\n<p style=\"text-align: center;\">[latex]\\begin{align} \\cos x\\cos \\left(2x\\right)+\\sin x\\sin \\left(2x\\right)&amp;=\\frac{\\sqrt{3}}{2} \\\\ \\cos \\left(x - 2x\\right)&amp;=\\frac{\\sqrt{3}}{2}&amp;&amp; \\text{Difference formula for cosine} \\\\ \\cos \\left(-x\\right)&amp;=\\frac{\\sqrt{3}}{2}&amp;&amp; \\text{Use the negative angle identity}. \\\\ \\cos x&amp;=\\frac{\\sqrt{3}}{2}\\end{align}[\/latex]<\/p>\r\nFrom the unit circle in Sum and Difference Identities, we see that [latex]\\cos x=\\frac{\\sqrt{3}}{2}[\/latex] when [latex]x=\\frac{\\pi }{6},\\frac{11\\pi }{6}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 15: Solving the Equation Using a Double-Angle Formula<\/h3>\r\nSolve the equation exactly using a double-angle formula: [latex]\\cos \\left(2\\theta \\right)=\\cos \\theta [\/latex].\r\n\r\n[reveal-answer q=\"631761\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"631761\"]\r\n\r\nWe 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:\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\cos \\left(2\\theta \\right)=\\cos \\theta \\\\ 2{\\cos }^{2}\\theta -1=\\cos \\theta \\\\ 2{\\cos }^{2}\\theta -\\cos \\theta -1=0 \\\\ \\left(2\\cos \\theta +1\\right)\\left(\\cos \\theta -1\\right)=0 \\\\ 2\\cos \\theta +1=0 \\\\ \\cos \\theta =-\\frac{1}{2} \\\\ \\text{ } \\\\ \\cos \\theta -1=0 \\\\ \\cos \\theta =1 \\end{gathered}[\/latex]<\/p>\r\nSo, if [latex]\\cos \\theta =-\\frac{1}{2}[\/latex], then [latex]\\theta =\\frac{2\\pi }{3}\\pm 2\\pi k[\/latex] and [latex]\\theta =\\frac{4\\pi }{3}\\pm 2\\pi k[\/latex]; if [latex]\\cos \\theta =1[\/latex], then [latex]\\theta =0\\pm 2\\pi k[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 16: Solving an Equation Using an Identity<\/h3>\r\nSolve the equation exactly using an identity: [latex]3\\cos \\theta +3=2{\\sin }^{2}\\theta ,0\\le \\theta &lt;2\\pi [\/latex].\r\n\r\n[reveal-answer q=\"468404\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"468404\"]\r\n\r\nIf we rewrite the right side, we can write the equation in terms of cosine:\r\n<p style=\"text-align: center;\">[latex]\\begin{align} 3 cos\\theta +3&amp; ={2 sin}^{2}\\theta \\\\ 3 cos\\theta +3&amp; =2\\left(1-{\\text{cos}}^{2}\\theta \\right) \\\\ 3 cos\\theta +3&amp; =2 - 2{\\cos }^{2}\\theta \\\\ 2{\\cos }^{2}\\theta +3 cos\\theta +1&amp; =0 \\\\ \\left(2 cos\\theta +1\\right)\\left(\\cos \\theta +1\\right)&amp; =0 \\\\ 2 cos\\theta +1&amp; =0 \\\\ \\cos \\theta &amp; =-\\frac{1}{2} \\\\ \\theta &amp; =\\frac{2\\pi }{3},\\frac{4\\pi }{3} \\\\ \\text{ } \\\\ \\cos \\theta +1&amp; =0 \\\\ \\cos \\theta &amp; =-1 \\\\ \\theta &amp; =\\pi\\end{align}[\/latex]<\/p>\r\nOur solutions are [latex]\\theta =\\frac{2\\pi }{3},\\frac{4\\pi }{3},\\pi [\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]149012[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Solving Trigonometric Equations with Multiple Angles<\/h2>\r\nSometimes it is not possible to solve a trigonometric equation with identities that have a multiple angle, such as [latex]\\sin \\left(2x\\right)[\/latex] or [latex]\\cos \\left(3x\\right)[\/latex]. When confronted with these equations, recall that [latex]y=\\sin \\left(2x\\right)[\/latex] is a <strong>horizontal compression<\/strong> by a factor of 2 of the function [latex]y=\\sin x[\/latex]. On an interval of [latex]2\\pi [\/latex], we can graph two periods of [latex]y=\\sin \\left(2x\\right)[\/latex], as opposed to one cycle of [latex]y=\\sin x[\/latex]. This compression of the graph leads us to believe there may be twice as many <em>x<\/em>-intercepts or solutions to [latex]\\sin \\left(2x\\right)=0[\/latex] compared to [latex]\\sin x=0[\/latex]. This information will help us solve the equation.\r\n<div class=\"textbox shaded\">\r\n<h3>Example 17: Solving a Multiple Angle Trigonometric Equation<\/h3>\r\nSolve exactly: [latex]\\cos \\left(2x\\right)=\\frac{1}{2}[\/latex] on [latex]\\left[0,2\\pi \\right)[\/latex].\r\n\r\n[reveal-answer q=\"299198\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"299198\"]\r\n\r\nWe can see that this equation is the standard equation with a multiple of an angle. If [latex]\\cos \\left(\\alpha \\right)=\\frac{1}{2}[\/latex], we know [latex]\\alpha [\/latex] is in quadrants I and IV. While [latex]\\theta ={\\cos }^{-1}\\frac{1}{2}[\/latex] will only yield solutions in quadrants I and II, we recognize that the solutions to the equation [latex]\\cos \\theta =\\frac{1}{2}[\/latex] will be in quadrants I and IV.\r\n\r\nTherefore, 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]. Does this make sense? Yes, because [latex]\\cos \\left(2\\left(\\frac{\\pi }{6}\\right)\\right)=\\cos \\left(\\frac{\\pi }{3}\\right)=\\frac{1}{2}[\/latex].\r\n\r\nAre there any other possible answers? Let us return to our first step.\r\n\r\nIn quadrant I, [latex]2x=\\frac{\\pi }{3}[\/latex], so [latex]x=\\frac{\\pi }{6}[\/latex] as noted. Let us revolve around the circle again:\r\n<p style=\"text-align: center;\">[latex]\\begin{align} 2x&amp;=\\frac{\\pi }{3}+2\\pi \\\\ &amp;=\\frac{\\pi }{3}+\\frac{6\\pi }{3} \\\\ &amp;=\\frac{7\\pi }{3} \\end{align}[\/latex]<\/p>\r\nso [latex]x=\\frac{7\\pi }{6}[\/latex].\r\n\r\nOne more rotation yields\r\n<p style=\"text-align: center;\">[latex]\\begin{align} 2x&amp;=\\frac{\\pi }{3}+4\\pi \\\\ &amp;=\\frac{\\pi }{3}+\\frac{12\\pi }{3} \\\\ &amp;=\\frac{13\\pi }{3} \\end{align}[\/latex]<\/p>\r\n[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].\r\n\r\nIn 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:\r\n<p style=\"text-align: center;\">[latex]\\begin{align}2x&amp;=\\frac{5\\pi }{3}+2\\pi \\\\ &amp;=\\frac{5\\pi }{3}+\\frac{6\\pi }{3} \\\\ &amp;=\\frac{11\\pi }{3}\\end{align}[\/latex]<\/p>\r\nso [latex]x=\\frac{11\\pi }{6}[\/latex].\r\n\r\nOne more rotation yields\r\n<p style=\"text-align: center;\">[latex]\\begin{align}2x&amp;=\\frac{5\\pi }{3}+4\\pi \\\\ &amp;=\\frac{5\\pi }{3}+\\frac{12\\pi }{3} \\\\ &amp;=\\frac{17\\pi }{3} \\end{align}[\/latex]<\/p>\r\n[latex]x=\\frac{17\\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].\r\n\r\nOur solutions are [latex]x=\\frac{\\pi }{6},\\frac{5\\pi }{6},\\frac{7\\pi }{6},\\text{and }\\frac{11\\pi }{6}[\/latex]. Note that whenever we solve a problem in the form of [latex]\\sin \\left(nx\\right)=c[\/latex], we must go around the unit circle [latex]n[\/latex] times.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Solving Right Triangle Problems<\/h2>\r\nWe can now use all of the methods we have learned to solve problems that involve applying the properties of right triangles and the <strong>Pythagorean Theorem<\/strong>. We begin with the familiar Pythagorean Theorem, [latex]{a}^{2}+{b}^{2}={c}^{2}[\/latex], and model an equation to fit a situation.\r\n<div class=\"textbox shaded\">\r\n<h3>Example 18: Using the Pythagorean Theorem to Model an Equation<\/h3>\r\nUse the Pythagorean Theorem, and the properties of right triangles to model an equation that fits the problem.\r\n\r\nOne of the cables that anchors the center of the London Eye Ferris wheel to the ground must be replaced. The center of the Ferris wheel is 69.5 meters above the ground, and the second anchor on the ground is 23 meters from the base of the Ferris wheel. Approximately how long is the cable, and what is the angle of elevation (from ground up to the center of the Ferris wheel)?\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164115\/CNX_Precalc_Figure_07_05_0022.jpg\" alt=\"Basic diagram of a ferris wheel (circle) and its support cables (form a right triangle). One cable runs from the center of the circle to the ground (outside the circle), is perpendicular to the ground, and has length 69.5. Another cable of unknown length (the hypotenuse) runs from the center of the circle to the ground 23 feet away from the other cable at an angle of theta degrees with the ground. So, in closing, there is a right triangle with base 23, height 69.5, hypotenuse unknown, and angle between base and hypotenuse of theta degrees.\" width=\"487\" height=\"278\" \/> <b>Figure 4<\/b>[\/caption]\r\n\r\n[reveal-answer q=\"808993\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"808993\"]\r\n\r\nUsing the information given, we can draw a right triangle. We can find the length of the cable with the Pythagorean Theorem.\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}{a}^{2}+{b}^{2}={c}^{2} \\\\ {\\left(23\\right)}^{2}+{\\left(69.5\\right)}^{2}\\approx 5359 \\\\ \\sqrt{5359}\\approx 73.2\\text{ m}\\end{gathered}[\/latex]<\/p>\r\nThe angle of elevation is [latex]\\theta [\/latex], formed by the second anchor on the ground and the cable reaching to the center of the wheel. We can use the tangent function to find its measure. Round to two decimal places.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}\\tan \\theta &amp;=\\frac{69.5}{23} \\\\ {\\tan }^{-1}\\left(\\frac{69.5}{23}\\right)&amp;\\approx 1.2522 \\\\ &amp;\\approx {71.69}^{\\circ } \\end{align}[\/latex]<\/p>\r\nThe angle of elevation is approximately [latex]{71.7}^{\\circ }[\/latex], and the length of the cable is 73.2 meters.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3>Example 19: Using the Pythagorean Theorem to Model an Abstract Problem<\/h3>\r\nOSHA safety regulations require that the base of a ladder be placed 1 foot from the wall for every 4 feet of ladder length. Find the angle that a ladder of any length forms with the ground and the height at which the ladder touches the wall.\r\n\r\n[reveal-answer q=\"335979\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"335979\"]\r\n\r\nFor any length of ladder, the base needs to be a distance from the wall equal to one fourth of the ladder\u2019s length. Equivalently, if the base of the ladder is \"<em>a\"<\/em> feet from the wall, the length of the ladder will be 4<em>a<\/em> feet.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164118\/CNX_Precalc_Figure_07_05_0032.jpg\" alt=\"Diagram of a right triangle with base length a, height length b, hypotenuse length 4a. Opposite the height is an angle of theta degrees, and opposite the hypotenuse is an angle of 90 degrees.\" width=\"487\" height=\"206\" \/> <b>Figure 5<\/b>[\/caption]\r\n\r\nThe side adjacent to [latex]\\theta [\/latex] is <em>a <\/em>and the hypotenuse is [latex]4a[\/latex]. Thus,\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\cos \\theta =\\frac{a}{4a}=\\frac{1}{4} \\\\ {\\cos }^{-1}\\left(\\frac{1}{4}\\right)\\approx {75.5}^{\\circ } \\end{gathered}[\/latex]<\/p>\r\nThe elevation of the ladder forms an angle of [latex]{75.5}^{\\circ }[\/latex] with the ground. The height at which the ladder touches the wall can be found using the Pythagorean Theorem:\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}{a}^{2}+{b}^{2}={\\left(4a\\right)}^{2} \\\\ {b}^{2}={\\left(4a\\right)}^{2}-{a}^{2} \\\\ {b}^{2}=16{a}^{2}-{a}^{2} \\\\ {b}^{2}=15{a}^{2} \\\\ b=\\sqrt{15}a \\end{gathered}[\/latex]<\/p>\r\nThus, the ladder touches the wall at [latex]\\sqrt{15}a[\/latex] feet from the ground.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h1>Key Concepts<\/h1>\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 \t<li>Real-world scenarios can be modeled and solved using the Pythagorean Theorem and trigonometric functions.<\/li>\r\n<\/ul>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li style=\"font-weight: 400;\">Solve equations involving a single trigonometric function.<\/li>\n<li style=\"font-weight: 400;\">Solve trigonometric equations using a calculator.<\/li>\n<li style=\"font-weight: 400;\">Solve trigonometric equations that involve factoring.<\/li>\n<li style=\"font-weight: 400;\">Solve trigonometric equations using fundamental identities.<\/li>\n<li style=\"font-weight: 400;\">Solve trigonometric equations with multiple angles.<\/li>\n<li style=\"font-weight: 400;\">Solve right triangle problems.<\/li>\n<\/ul>\n<\/div>\n<h2>Solving Linear Trigonometric Equations in Sine and Cosine<\/h2>\n<p>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. 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. Additionally, like rational equations, the domain of the function must be considered before we assume that any solution is valid. The <strong>period<\/strong> 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. 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. Recall the rule that gives the format for stating all possible solutions for a function where the period is [latex]2\\pi :[\/latex]<\/p>\n<div style=\"text-align: center;\">[latex]\\sin \\theta =\\sin \\left(\\theta \\pm 2k\\pi \\right)[\/latex]<\/div>\n<p>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 class=\"textbox shaded\">\n<h3>Example 1: Solving a Linear Trigonometric Equation Involving the Cosine Function<\/h3>\n<p>Find all possible exact solutions for the equation [latex]\\cos \\theta =\\frac{1}{2}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q84784\">Show Solution<\/span><\/p>\n<div id=\"q84784\" class=\"hidden-answer\" style=\"display: none\">\n<p>From the <strong>unit circle<\/strong>, we know that<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\cos \\theta =\\frac{1}{2} \\\\ \\theta =\\frac{\\pi }{3},\\frac{5\\pi }{3} \\end{gathered}[\/latex]<\/p>\n<p>These are the solutions in the interval [latex]\\left[0,2\\pi \\right][\/latex]. All possible solutions are given by<\/p>\n<p style=\"text-align: center;\">[latex]\\theta =\\frac{\\pi }{3}\\pm 2k\\pi \\text{ and }\\theta =\\frac{5\\pi }{3}\\pm 2k\\pi[\/latex]<\/p>\n<p>where [latex]k[\/latex] is an integer.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 2: Solving a Linear Equation Involving the Sine Function<\/h3>\n<p>Find all possible exact solutions for the equation [latex]\\sin t=\\frac{1}{2}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q535703\">Show Solution<\/span><\/p>\n<div id=\"q535703\" class=\"hidden-answer\" style=\"display: none\">\n<p>Solving for all possible values of <em>t<\/em> means that solutions include angles beyond the period of [latex]2\\pi[\/latex]. From the unit circle, we can see that the solutions are [latex]t=\\frac{\\pi }{6}[\/latex] and [latex]t=\\frac{5\\pi }{6}[\/latex]. But the problem is asking for all possible values that solve the equation. Therefore, the answer is<\/p>\n<p style=\"text-align: center;\">[latex]t=\\frac{\\pi }{6}\\pm 2\\pi k\\text{ and }t=\\frac{5\\pi }{6}\\pm 2\\pi k[\/latex]<\/p>\n<p>where [latex]k[\/latex] is an integer.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a trigonometric equation, solve using algebra.<\/h3>\n<ul>\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<\/ul>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 3: Solve the Trigonometric Equation in Linear Form<\/h3>\n<p>Solve the equation exactly: [latex]2\\cos \\theta -3=-5,0\\le \\theta <2\\pi[\/latex].\n\n\n\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q939405\">Show Solution<\/span><\/p>\n<p style=\"text-align: left;\">\n<div id=\"q939405\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{gathered}2\\cos \\theta -3=-5 \\\\ \\cos \\theta =-2 \\\\ \\cos \\theta =-1 \\\\ \\theta =\\pi \\end{gathered}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Solve exactly the following linear equation on the interval [latex]\\left[0,2\\pi \\right):2\\sin x+1=0[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q500341\">Show Solution<\/span><\/p>\n<div id=\"q500341\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x=\\frac{7\\pi }{6},\\frac{11\\pi }{6}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm149871\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=149871&theme=oea&iframe_resize_id=ohm149871\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Solve Trigonometric Equations Using a Calculator<\/h2>\n<p>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 class=\"textbox shaded\">\n<h3>Example 4: Using a Calculator to Solve a Trigonometric Equation Involving Sine<\/h3>\n<p>Use a calculator to solve the equation [latex]\\sin \\theta =0.8[\/latex], where [latex]\\theta[\/latex] is in radians.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q522194\">Show Solution<\/span><\/p>\n<div id=\"q522194\" class=\"hidden-answer\" style=\"display: none\">\n<h3><\/h3>\n<p>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]{\\sin }^{-1}[\/latex] function. What is shown on the screen is [latex]{\\sin}^{-1}[\/latex](. The calculator is ready for the input within the parentheses. For this problem, we enter [latex]{\\sin }^{-1}\\left(0.8\\right)[\/latex], and press ENTER. Thus, to four decimals places,<\/p>\n<p style=\"text-align: center;\">[latex]{\\sin }^{-1}\\left(0.8\\right)\\approx 0.9273[\/latex]<\/p>\n<p>This is the solution in quadrant I. There is also a solution in quadrant II. To find this we subtract [latex]\/pi - 0.9273 \\approx 2.2143[\/latex]<\/p>\n<p>The general solution is<\/p>\n<p style=\"text-align: center;\">[latex]\\theta \\approx 0.9273\\pm 2\\pi k \\text{ and } \\theta \\approx 2.2143 \\pm 2\\pi k[\/latex]<\/p>\n<p>The angle measurement in degrees is<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} \\theta &\\approx {53.1}^{\\circ } \\\\ \\theta &\\approx {180}^{\\circ }-{53.1}^{\\circ } \\\\ &\\approx {126.9}^{\\circ } \\end{align}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>Note that a calculator will only return an angle in quadrants I or IV for the sine function, since that is the range of the inverse sine. The other angle is obtained by using [latex]\\pi -\\theta[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 5: Using a Calculator to Solve a Trigonometric Equation Involving Secant<\/h3>\n<p>Use a calculator to solve the equation [latex]\\sec \\theta =-4[\/latex], giving your answer in radians.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q209133\">Show Solution<\/span><\/p>\n<div id=\"q209133\" class=\"hidden-answer\" style=\"display: none\">\n<p>We can begin with some algebra.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\sec \\theta =-4\\\\ \\frac{1}{\\cos \\theta }=-4\\\\ \\cos \\theta =-\\frac{1}{4}\\end{gathered}[\/latex]<\/p>\n<p>Check that the MODE is in radians. Now use the inverse cosine function.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}{\\cos }^{-1}\\left(-\\frac{1}{4}\\right)\\approx 1.8235 \\\\ \\theta \\approx 1.8235+2\\pi k \\end{gathered}[\/latex]<\/p>\n<p>Since [latex]\\frac{\\pi }{2}\\approx 1.57[\/latex] and [latex]\\pi \\approx 3.14[\/latex], 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.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164111\/CNX_Precalc_Figure_07_05_0052.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=\"487\" height=\"380\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 2.<\/b><\/p>\n<\/div>\n<p>So, we also need to find the measure of the angle in quadrant III. In quadrant III, the reference angle is [latex]\\theta \\text{ }\\text{ }\\text{'}\\approx \\pi -\\text{1}\\text{.8235}\\approx \\text{1}\\text{.3181}\\text{.}[\/latex] The other solution in quadrant III is [latex]\\theta \\text{ }\\text{ }\\text{'}\\approx \\pi +\\text{1}\\text{.3181}\\approx \\text{4}\\text{.4597}\\text{.}[\/latex]<\/p>\n<p>The solutions are [latex]\\theta \\approx 1.8235\\pm 2\\pi k[\/latex] and [latex]\\theta \\approx 4.4597\\pm 2\\pi k[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Solve [latex]\\cos \\theta =-0.2[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q145806\">Show Solution<\/span><\/p>\n<div id=\"q145806\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\theta \\approx 1.7722\\pm 2\\pi k[\/latex] and [latex]\\theta \\approx 4.5110\\pm 2\\pi k[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm149873\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=149873&theme=oea&iframe_resize_id=ohm149873\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Solving Equations Involving a Single Trigonometric Function<\/h2>\n<p>When we are given equations that involve only one of the six trigonometric functions, their solutions involve using algebraic techniques and 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 reciprocal function, and solve for the angles using the function. 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 class=\"textbox shaded\">\n<h3>Example 6: Solving a Problem Involving a Single Trigonometric Function<\/h3>\n<p>Solve the problem exactly: [latex]2{\\sin }^{2}\\theta -1=0,0\\le \\theta <2\\pi[\/latex].\n\n\n\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q467313\">Show Solution<\/span><\/p>\n<div id=\"q467313\" class=\"hidden-answer\" style=\"display: none\">\n<p>As this problem is not easily factored, we will solve using the square root property. First, we use algebra to isolate [latex]\\sin \\theta[\/latex]. Then we will find the angles.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}2{\\sin }^{2}\\theta -1=0 \\\\ 2{\\sin }^{2}\\theta =1 \\\\ {\\sin }^{2}\\theta =\\frac{1}{2} \\\\ \\sqrt{{\\sin }^{2}\\theta }=\\pm \\sqrt{\\frac{1}{2}} \\\\ \\sin \\theta =\\pm \\frac{1}{\\sqrt{2}}=\\pm \\frac{\\sqrt{2}}{2} \\\\ \\theta =\\frac{\\pi }{4},\\frac{3\\pi }{4},\\frac{5\\pi }{4},\\frac{7\\pi }{4} \\end{gathered}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 7: Solving a Trigonometric Equation Involving Cosecant<\/h3>\n<p>Solve the following equation exactly: [latex]\\csc \\theta =-2,0\\le \\theta <4\\pi[\/latex].\n\n\n\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q605306\">Show Solution<\/span><\/p>\n<div id=\"q605306\" class=\"hidden-answer\" style=\"display: none\">\n<p>We want all values of [latex]\\theta[\/latex] for which [latex]\\csc \\theta =-2[\/latex] over the interval [latex]0\\le \\theta <4\\pi[\/latex].\n\n\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\csc \\theta =-2 \\\\ \\frac{1}{\\sin \\theta }=-2 \\\\ \\sin \\theta =-\\frac{1}{2} \\\\ \\theta =\\frac{7\\pi }{6},\\frac{11\\pi }{6},\\frac{19\\pi }{6},\\frac{23\\pi }{6} \\end{gathered}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>As [latex]\\sin \\theta =-\\frac{1}{2}[\/latex], notice that all four solutions are in the third and fourth quadrants.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 8: Solving an Equation Involving Tangent<\/h3>\n<p>Solve the equation exactly: [latex]\\tan \\left(\\theta -\\frac{\\pi }{2}\\right)=1,0\\le \\theta <2\\pi[\/latex].\n\n\n\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q484899\">Show Solution<\/span><\/p>\n<div id=\"q484899\" class=\"hidden-answer\" style=\"display: none\">\n<p>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]\\tan \\left(\\frac{\\pi }{4}\\right)=1[\/latex], then<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\theta -\\frac{\\pi }{2}=\\frac{\\pi }{4}\\\\ \\theta =\\frac{3\\pi }{4}\\pm k\\pi \\end{gathered}[\/latex]<\/p>\n<p>Over the interval [latex]\\left[0,2\\pi \\right)[\/latex], we have two solutions:<\/p>\n<p style=\"text-align: center;\">[latex]\\theta =\\frac{3\\pi }{4}\\text{ and }\\theta =\\frac{3\\pi }{4}+\\pi =\\frac{7\\pi }{4}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Find all solutions for [latex]\\tan x=\\sqrt{3}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q629684\">Show Solution<\/span><\/p>\n<div id=\"q629684\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{\\pi }{3}\\pm \\pi k[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm173739\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=173739&theme=oea&iframe_resize_id=ohm173739\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 9: Identify all Solutions to the Equation Involving Tangent<\/h3>\n<p>Identify all exact solutions to the equation [latex]2\\left(\\tan x+3\\right)=5+\\tan x,0\\le x<2\\pi[\/latex].\n\n\n\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q949694\">Show Solution<\/span><\/p>\n<div id=\"q949694\" class=\"hidden-answer\" style=\"display: none\">\n<p>We can solve this equation using only algebra. Isolate the expression [latex]\\tan x[\/latex] on the left side of the equals sign.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered} 2\\left(\\tan x\\right)+2\\left(3\\right) =5+\\tan x \\\\ 2\\tan x+6 =5+\\tan x \\\\ 2\\tan x-\\tan x =5 - 6 \\\\ \\tan x =-1\\end{gathered}[\/latex]<\/p>\n<p>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<h2>Solving Trigonometric Equations in Quadratic Form<\/h2>\n<p>Solving a <strong>quadratic equation<\/strong> 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 shaded\">\n<h3>Example 10: Solving a Trigonometric Equation in Quadratic Form<\/h3>\n<p>Solve the equation exactly: [latex]{\\cos }^{2}\\theta +3\\cos \\theta -1=0,0\\le \\theta <2\\pi[\/latex].\n\n\n\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q563509\">Show Solution<\/span><\/p>\n<div id=\"q563509\" class=\"hidden-answer\" style=\"display: none\">\n<p>We begin by using substitution and replacing cos [latex]\\theta[\/latex] with [latex]x[\/latex]. It is not necessary to use substitution, but it may make the problem easier to solve visually. Let [latex]\\cos \\theta =x[\/latex]. We have<\/p>\n<p style=\"text-align: center;\">[latex]{x}^{2}+3x - 1=0[\/latex]<\/p>\n<p>The equation cannot be factored, so we will use the <strong>quadratic formula<\/strong> [latex]x=\\frac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a}[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} x&=\\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]<\/p>\n<p>Replace [latex]x[\/latex] with [latex]\\cos \\theta[\/latex], and solve. Thus,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered} \\cos \\theta =\\frac{-3\\pm \\sqrt{13}}{2}\\theta ={\\cos }^{-1}\\left(\\frac{-3+\\sqrt{13}}{2}\\right)\\end{gathered}[\/latex]<\/p>\n<p>Note that only the + sign is used. This is because we get an error when we solve [latex]\\theta ={\\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]. However, there is a second solution:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\theta &={\\cos }^{-1}\\left(\\frac{-3+\\sqrt{13}}{2}\\right) \\\\ &\\approx 1.26 \\end{align}[\/latex]<\/p>\n<p>This terminal side of the angle lies in quadrant I. Since cosine is also positive in quadrant IV, the second solution is<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\theta &=2\\pi -{\\cos }^{-1}\\left(\\frac{-3+\\sqrt{13}}{2}\\right) \\\\ &\\approx 5.02 \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 11: Solving a Trigonometric Equation in Quadratic Form by Factoring<\/h3>\n<p>Solve the equation exactly: [latex]2{\\sin }^{2}\\theta -5\\sin \\theta +3=0,0\\le \\theta \\le 2\\pi[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q693370\">Show Solution<\/span><\/p>\n<div id=\"q693370\" class=\"hidden-answer\" style=\"display: none\">\n<p>Using grouping, this quadratic can be factored. Either make the real substitution, [latex]\\sin \\theta =u[\/latex], or imagine it, as we factor:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}2{\\sin }^{2}\\theta -5\\sin \\theta +3=0 \\\\ \\left(2\\sin \\theta -3\\right)\\left(\\sin \\theta -1\\right)=0 \\end{gathered}[\/latex]<\/p>\n<p>Now set each factor equal to zero.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}2\\sin \\theta -3=0 \\\\ 2\\sin \\theta =3 \\\\ \\sin \\theta =\\frac{3}{2} \\\\ \\text{ } \\\\ \\sin \\theta -1=0 \\\\ \\sin \\theta =1 \\end{gathered}[\/latex]<\/p>\n<p>Next solve for [latex]\\theta :\\sin \\theta \\ne \\frac{3}{2}[\/latex], as the range of the sine function is [latex]\\left[-1,1\\right][\/latex]. However, [latex]\\sin \\theta =1[\/latex], giving the solution [latex]\\theta =\\frac{\\pi }{2}[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>Make sure to check all solutions on the given domain as some factors have no solution.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Solve [latex]{\\sin }^{2}\\theta =2\\cos \\theta +2,0\\le \\theta \\le 2\\pi[\/latex]. [Hint: Make a substitution to express the equation only in terms of cosine.]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q608693\">Show Solution<\/span><\/p>\n<div id=\"q608693\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\cos \\theta =-1,\\theta =\\pi[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm149010\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=149010&theme=oea&iframe_resize_id=ohm149010\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 12: Solving a Trigonometric Equation Using Algebra<\/h3>\n<p>Solve exactly:<\/p>\n<p style=\"text-align: center;\">[latex]2{\\sin }^{2}\\theta +\\sin \\theta =0;0\\le \\theta <2\\pi[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q204542\">Show Solution<\/span><\/p>\n<div id=\"q204542\" class=\"hidden-answer\" style=\"display: none\">\n<p>This problem should appear familiar as it is similar to a quadratic. Let [latex]\\sin \\theta =x[\/latex]. The equation becomes [latex]2{x}^{2}+x=0[\/latex]. We begin by factoring:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}2{x}^{2}+x=0\\\\ x\\left(2x+1\\right)=0\\end{gathered}[\/latex]<\/p>\n<p>Set each factor equal to zero.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}x=0 \\\\ 2x+1=0 \\\\ x=-\\frac{1}{2} \\end{gathered}[\/latex]<\/p>\n<p>Then, substitute back into the equation the original expression [latex]\\sin \\theta[\/latex] for [latex]x[\/latex]. Thus,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\sin \\theta =0 \\\\ \\theta =0,\\pi \\\\ \\text{ } \\\\ \\sin \\theta =-\\frac{1}{2} \\\\ \\theta =\\frac{7\\pi }{6},\\frac{11\\pi }{6} \\end{gathered}[\/latex]<\/p>\n<p>The solutions within the domain [latex]0\\le \\theta <2\\pi[\/latex] are [latex]\\theta =0,\\pi ,\\frac{7\\pi }{6},\\frac{11\\pi }{6}[\/latex].\n\nIf we prefer not to substitute, we can solve the equation by following the same pattern of factoring and setting each factor equal to zero.\n\n\n<p style=\"text-align: center;\">[latex]\\begin{gathered}2{\\sin }^{2}\\theta +\\sin \\theta =0 \\\\ \\sin \\theta \\left(2\\sin \\theta +1\\right)=0 \\\\ \\sin \\theta =0 \\\\ \\theta =0,\\pi \\\\ \\text{ } \\\\ 2\\sin \\theta +1=0 \\\\ 2\\sin \\theta =-1 \\\\ \\sin \\theta =-\\frac{1}{2} \\\\ \\theta =\\frac{7\\pi }{6},\\frac{11\\pi }{6} \\end{gathered}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>We can see the solutions on the graph in Figure 3. On the interval [latex]0\\le \\theta <2\\pi[\/latex], the graph crosses the <em>x-<\/em>axis four times, at the solutions noted. Notice that trigonometric equations that are in quadratic form can yield up to four solutions instead of the expected two that are found with quadratic equations. In this example, each solution (angle) corresponding to a positive sine value will yield two angles that would result in that value.<\/p>\n<div style=\"width: 741px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164113\/CNX_Precalc_Figure_07_05_0042.jpg\" alt=\"Graph of 2*(sin(theta))^2 + sin(theta) from 0 to 2pi. Zeros are at 0, pi, 7pi\/6, and 11pi\/6.\" width=\"731\" height=\"256\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 3<\/b><\/p>\n<\/div>\n<p>We can verify the solutions on the <a href=\"https:\/\/courses.lumenlearning.com\/precalctwoxmaster\/chapter\/use-sum-and-difference-formulas-for-cosine\/\" target=\"_blank\" rel=\"noopener\">unit circle<\/a> in Sum and Difference Identities\u00a0as well.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 13: Solving a Trigonometric Equation Quadratic in Form<\/h3>\n<p>Solve the equation quadratic in form exactly: [latex]2{\\sin }^{2}\\theta -3\\sin \\theta +1=0,0\\le \\theta <2\\pi[\/latex].\n\n\n\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q699307\">Show Solution<\/span><\/p>\n<div id=\"q699307\" class=\"hidden-answer\" style=\"display: none\">\n<p>We can factor using grouping. Solution values of [latex]\\theta[\/latex] can be found on the unit circle:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\left(2\\sin \\theta -1\\right)\\left(\\sin \\theta -1\\right)=0 \\\\ 2\\sin \\theta -1=0 \\\\ \\sin \\theta =\\frac{1}{2} \\\\ \\theta =\\frac{\\pi }{6},\\frac{5\\pi }{6} \\\\ \\text{ } \\\\ \\sin \\theta =1 \\\\ \\theta =\\frac{\\pi }{2} \\end{gathered}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p>Solve the quadratic equation [latex]2{\\cos }^{2}\\theta +\\cos \\theta =0[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q736189\">Show Solution<\/span><\/p>\n<div id=\"q736189\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac{\\pi }{2},\\frac{2\\pi }{3},\\frac{4\\pi }{3},\\frac{3\\pi }{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Solving Trigonometric Equations Using Fundamental Identities<\/h2>\n<p>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<div class=\"textbox shaded\">\n<h3>Example 14: Use Identities to Solve an Equation<\/h3>\n<p>Use identities to solve exactly the trigonometric equation over the interval [latex]0\\le x<2\\pi[\/latex].\n\n\n<p style=\"text-align: center;\">[latex]\\cos x\\cos \\left(2x\\right)+\\sin x\\sin \\left(2x\\right)=\\frac{\\sqrt{3}}{2}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q607762\">Show Solution<\/span><\/p>\n<div id=\"q607762\" class=\"hidden-answer\" style=\"display: none\">\n<p>Notice that the left side of the equation is the difference formula for cosine.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} \\cos x\\cos \\left(2x\\right)+\\sin x\\sin \\left(2x\\right)&=\\frac{\\sqrt{3}}{2} \\\\ \\cos \\left(x - 2x\\right)&=\\frac{\\sqrt{3}}{2}&& \\text{Difference formula for cosine} \\\\ \\cos \\left(-x\\right)&=\\frac{\\sqrt{3}}{2}&& \\text{Use the negative angle identity}. \\\\ \\cos x&=\\frac{\\sqrt{3}}{2}\\end{align}[\/latex]<\/p>\n<p>From the unit circle in Sum and Difference Identities, we see that [latex]\\cos x=\\frac{\\sqrt{3}}{2}[\/latex] when [latex]x=\\frac{\\pi }{6},\\frac{11\\pi }{6}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 15: Solving the Equation Using a Double-Angle Formula<\/h3>\n<p>Solve the equation exactly using a double-angle formula: [latex]\\cos \\left(2\\theta \\right)=\\cos \\theta[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q631761\">Show Solution<\/span><\/p>\n<div id=\"q631761\" class=\"hidden-answer\" style=\"display: none\">\n<p>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<p style=\"text-align: center;\">[latex]\\begin{gathered}\\cos \\left(2\\theta \\right)=\\cos \\theta \\\\ 2{\\cos }^{2}\\theta -1=\\cos \\theta \\\\ 2{\\cos }^{2}\\theta -\\cos \\theta -1=0 \\\\ \\left(2\\cos \\theta +1\\right)\\left(\\cos \\theta -1\\right)=0 \\\\ 2\\cos \\theta +1=0 \\\\ \\cos \\theta =-\\frac{1}{2} \\\\ \\text{ } \\\\ \\cos \\theta -1=0 \\\\ \\cos \\theta =1 \\end{gathered}[\/latex]<\/p>\n<p>So, if [latex]\\cos \\theta =-\\frac{1}{2}[\/latex], then [latex]\\theta =\\frac{2\\pi }{3}\\pm 2\\pi k[\/latex] and [latex]\\theta =\\frac{4\\pi }{3}\\pm 2\\pi k[\/latex]; if [latex]\\cos \\theta =1[\/latex], then [latex]\\theta =0\\pm 2\\pi k[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 16: Solving an Equation Using an Identity<\/h3>\n<p>Solve the equation exactly using an identity: [latex]3\\cos \\theta +3=2{\\sin }^{2}\\theta ,0\\le \\theta <2\\pi[\/latex].\n\n\n\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q468404\">Show Solution<\/span><\/p>\n<div id=\"q468404\" class=\"hidden-answer\" style=\"display: none\">\n<p>If we rewrite the right side, we can write the equation in terms of cosine:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} 3 cos\\theta +3& ={2 sin}^{2}\\theta \\\\ 3 cos\\theta +3& =2\\left(1-{\\text{cos}}^{2}\\theta \\right) \\\\ 3 cos\\theta +3& =2 - 2{\\cos }^{2}\\theta \\\\ 2{\\cos }^{2}\\theta +3 cos\\theta +1& =0 \\\\ \\left(2 cos\\theta +1\\right)\\left(\\cos \\theta +1\\right)& =0 \\\\ 2 cos\\theta +1& =0 \\\\ \\cos \\theta & =-\\frac{1}{2} \\\\ \\theta & =\\frac{2\\pi }{3},\\frac{4\\pi }{3} \\\\ \\text{ } \\\\ \\cos \\theta +1& =0 \\\\ \\cos \\theta & =-1 \\\\ \\theta & =\\pi\\end{align}[\/latex]<\/p>\n<p>Our solutions are [latex]\\theta =\\frac{2\\pi }{3},\\frac{4\\pi }{3},\\pi[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm149012\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=149012&theme=oea&iframe_resize_id=ohm149012\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Solving Trigonometric Equations with Multiple Angles<\/h2>\n<p>Sometimes it is not possible to solve a trigonometric equation with identities that have a multiple angle, such as [latex]\\sin \\left(2x\\right)[\/latex] or [latex]\\cos \\left(3x\\right)[\/latex]. When confronted with these equations, recall that [latex]y=\\sin \\left(2x\\right)[\/latex] is a <strong>horizontal compression<\/strong> by a factor of 2 of the function [latex]y=\\sin x[\/latex]. On an interval of [latex]2\\pi[\/latex], we can graph two periods of [latex]y=\\sin \\left(2x\\right)[\/latex], as opposed to one cycle of [latex]y=\\sin x[\/latex]. This compression of the graph leads us to believe there may be twice as many <em>x<\/em>-intercepts or solutions to [latex]\\sin \\left(2x\\right)=0[\/latex] compared to [latex]\\sin x=0[\/latex]. This information will help us solve the equation.<\/p>\n<div class=\"textbox shaded\">\n<h3>Example 17: Solving a Multiple Angle Trigonometric Equation<\/h3>\n<p>Solve exactly: [latex]\\cos \\left(2x\\right)=\\frac{1}{2}[\/latex] on [latex]\\left[0,2\\pi \\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q299198\">Show Solution<\/span><\/p>\n<div id=\"q299198\" class=\"hidden-answer\" style=\"display: none\">\n<p>We can see that this equation is the standard equation with a multiple of an angle. If [latex]\\cos \\left(\\alpha \\right)=\\frac{1}{2}[\/latex], we know [latex]\\alpha[\/latex] is in quadrants I and IV. While [latex]\\theta ={\\cos }^{-1}\\frac{1}{2}[\/latex] will only yield solutions in quadrants I and II, we recognize that the solutions to the equation [latex]\\cos \\theta =\\frac{1}{2}[\/latex] will be in quadrants I and IV.<\/p>\n<p>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]. Does this make sense? Yes, because [latex]\\cos \\left(2\\left(\\frac{\\pi }{6}\\right)\\right)=\\cos \\left(\\frac{\\pi }{3}\\right)=\\frac{1}{2}[\/latex].<\/p>\n<p>Are there any other possible answers? Let us return to our first step.<\/p>\n<p>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<p style=\"text-align: center;\">[latex]\\begin{align} 2x&=\\frac{\\pi }{3}+2\\pi \\\\ &=\\frac{\\pi }{3}+\\frac{6\\pi }{3} \\\\ &=\\frac{7\\pi }{3} \\end{align}[\/latex]<\/p>\n<p>so [latex]x=\\frac{7\\pi }{6}[\/latex].<\/p>\n<p>One more rotation yields<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} 2x&=\\frac{\\pi }{3}+4\\pi \\\\ &=\\frac{\\pi }{3}+\\frac{12\\pi }{3} \\\\ &=\\frac{13\\pi }{3} \\end{align}[\/latex]<\/p>\n<p>[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>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<p style=\"text-align: center;\">[latex]\\begin{align}2x&=\\frac{5\\pi }{3}+2\\pi \\\\ &=\\frac{5\\pi }{3}+\\frac{6\\pi }{3} \\\\ &=\\frac{11\\pi }{3}\\end{align}[\/latex]<\/p>\n<p>so [latex]x=\\frac{11\\pi }{6}[\/latex].<\/p>\n<p>One more rotation yields<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}2x&=\\frac{5\\pi }{3}+4\\pi \\\\ &=\\frac{5\\pi }{3}+\\frac{12\\pi }{3} \\\\ &=\\frac{17\\pi }{3} \\end{align}[\/latex]<\/p>\n<p>[latex]x=\\frac{17\\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>Our solutions are [latex]x=\\frac{\\pi }{6},\\frac{5\\pi }{6},\\frac{7\\pi }{6},\\text{and }\\frac{11\\pi }{6}[\/latex]. Note that whenever we solve a problem in the form of [latex]\\sin \\left(nx\\right)=c[\/latex], we must go around the unit circle [latex]n[\/latex] times.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Solving Right Triangle Problems<\/h2>\n<p>We can now use all of the methods we have learned to solve problems that involve applying the properties of right triangles and the <strong>Pythagorean Theorem<\/strong>. We begin with the familiar Pythagorean Theorem, [latex]{a}^{2}+{b}^{2}={c}^{2}[\/latex], and model an equation to fit a situation.<\/p>\n<div class=\"textbox shaded\">\n<h3>Example 18: Using the Pythagorean Theorem to Model an Equation<\/h3>\n<p>Use the Pythagorean Theorem, and the properties of right triangles to model an equation that fits the problem.<\/p>\n<p>One of the cables that anchors the center of the London Eye Ferris wheel to the ground must be replaced. The center of the Ferris wheel is 69.5 meters above the ground, and the second anchor on the ground is 23 meters from the base of the Ferris wheel. Approximately how long is the cable, and what is the angle of elevation (from ground up to the center of the Ferris wheel)?<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164115\/CNX_Precalc_Figure_07_05_0022.jpg\" alt=\"Basic diagram of a ferris wheel (circle) and its support cables (form a right triangle). One cable runs from the center of the circle to the ground (outside the circle), is perpendicular to the ground, and has length 69.5. Another cable of unknown length (the hypotenuse) runs from the center of the circle to the ground 23 feet away from the other cable at an angle of theta degrees with the ground. So, in closing, there is a right triangle with base 23, height 69.5, hypotenuse unknown, and angle between base and hypotenuse of theta degrees.\" width=\"487\" height=\"278\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 4<\/b><\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q808993\">Show Solution<\/span><\/p>\n<div id=\"q808993\" class=\"hidden-answer\" style=\"display: none\">\n<p>Using the information given, we can draw a right triangle. We can find the length of the cable with the Pythagorean Theorem.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}{a}^{2}+{b}^{2}={c}^{2} \\\\ {\\left(23\\right)}^{2}+{\\left(69.5\\right)}^{2}\\approx 5359 \\\\ \\sqrt{5359}\\approx 73.2\\text{ m}\\end{gathered}[\/latex]<\/p>\n<p>The angle of elevation is [latex]\\theta[\/latex], formed by the second anchor on the ground and the cable reaching to the center of the wheel. We can use the tangent function to find its measure. Round to two decimal places.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}\\tan \\theta &=\\frac{69.5}{23} \\\\ {\\tan }^{-1}\\left(\\frac{69.5}{23}\\right)&\\approx 1.2522 \\\\ &\\approx {71.69}^{\\circ } \\end{align}[\/latex]<\/p>\n<p>The angle of elevation is approximately [latex]{71.7}^{\\circ }[\/latex], and the length of the cable is 73.2 meters.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 19: Using the Pythagorean Theorem to Model an Abstract Problem<\/h3>\n<p>OSHA safety regulations require that the base of a ladder be placed 1 foot from the wall for every 4 feet of ladder length. Find the angle that a ladder of any length forms with the ground and the height at which the ladder touches the wall.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q335979\">Show Solution<\/span><\/p>\n<div id=\"q335979\" class=\"hidden-answer\" style=\"display: none\">\n<p>For any length of ladder, the base needs to be a distance from the wall equal to one fourth of the ladder\u2019s length. Equivalently, if the base of the ladder is &#8220;<em>a&#8221;<\/em> feet from the wall, the length of the ladder will be 4<em>a<\/em> feet.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27164118\/CNX_Precalc_Figure_07_05_0032.jpg\" alt=\"Diagram of a right triangle with base length a, height length b, hypotenuse length 4a. Opposite the height is an angle of theta degrees, and opposite the hypotenuse is an angle of 90 degrees.\" width=\"487\" height=\"206\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 5<\/b><\/p>\n<\/div>\n<p>The side adjacent to [latex]\\theta[\/latex] is <em>a <\/em>and the hypotenuse is [latex]4a[\/latex]. Thus,<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}\\cos \\theta =\\frac{a}{4a}=\\frac{1}{4} \\\\ {\\cos }^{-1}\\left(\\frac{1}{4}\\right)\\approx {75.5}^{\\circ } \\end{gathered}[\/latex]<\/p>\n<p>The elevation of the ladder forms an angle of [latex]{75.5}^{\\circ }[\/latex] with the ground. The height at which the ladder touches the wall can be found using the Pythagorean Theorem:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}{a}^{2}+{b}^{2}={\\left(4a\\right)}^{2} \\\\ {b}^{2}={\\left(4a\\right)}^{2}-{a}^{2} \\\\ {b}^{2}=16{a}^{2}-{a}^{2} \\\\ {b}^{2}=15{a}^{2} \\\\ b=\\sqrt{15}a \\end{gathered}[\/latex]<\/p>\n<p>Thus, the ladder touches the wall at [latex]\\sqrt{15}a[\/latex] feet from the ground.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h1>Key Concepts<\/h1>\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<li>Real-world scenarios can be modeled and solved using the Pythagorean Theorem and trigonometric functions.<\/li>\n<\/ul>\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-196\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: OpenStax College. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface<\/a>. <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":23485,"menu_order":10,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Precalculus\",\"author\":\"OpenStax College\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\",\"project\":\"\",\"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-196","chapter","type-chapter","status-publish","hentry"],"part":191,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-precalculus\/wp-json\/pressbooks\/v2\/chapters\/196","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-precalculus\/wp-json\/wp\/v2\/users\/23485"}],"version-history":[{"count":4,"href":"https:\/\/courses.lumenlearning.com\/gsu-precalculus\/wp-json\/pressbooks\/v2\/chapters\/196\/revisions"}],"predecessor-version":[{"id":673,"href":"https:\/\/courses.lumenlearning.com\/gsu-precalculus\/wp-json\/pressbooks\/v2\/chapters\/196\/revisions\/673"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-precalculus\/wp-json\/pressbooks\/v2\/parts\/191"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-precalculus\/wp-json\/pressbooks\/v2\/chapters\/196\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-precalculus\/wp-json\/wp\/v2\/media?parent=196"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=196"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-precalculus\/wp-json\/wp\/v2\/contributor?post=196"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-precalculus\/wp-json\/wp\/v2\/license?post=196"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}