{"id":3894,"date":"2021-05-20T18:29:55","date_gmt":"2021-05-20T18:29:55","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/review-for-maxima-and-minima\/"},"modified":"2021-07-03T19:03:25","modified_gmt":"2021-07-03T19:03:25","slug":"review-for-maxima-and-minima","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/review-for-maxima-and-minima\/","title":{"raw":"Skills Review for Maxima and Minima and The Mean Value Theorem","rendered":"Skills Review for Maxima and Minima and The Mean Value Theorem"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li><span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;Find the domain of a function algebraically&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4609,&quot;3&quot;:{&quot;1&quot;:0},&quot;12&quot;:0,&quot;15&quot;:&quot;Work Sans&quot;}\">Find the domain of a function algebraically<\/span><\/li>\r\n \t<li>Use the zero product principle to solve quadratic equations that can be factored<\/li>\r\n \t<li>Solve rational equations set equal to zero<\/li>\r\n \t<li>Solve for a variable in a square root equation<\/li>\r\n \t<li>Solve trigonometric equations<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn the Maxima and Minima and the Mean Value Theorem sections, we will find the domain of functions and their derivatives. Derivatives will also be set equal to 0 or some specific value in both sections which will lead to having to solve various types of equations. Here we will review how to find the domain of functions and how to solve various types of equations.\r\n<h2>Find the Domain of a Function<\/h2>\r\nRecall that the domain of a function is all input values where the function is defined. When given a function equation, you must remember the following three cases:\r\n<ol>\r\n \t<li>If the function has no denominator or an even root, in most cases, the domain will be all real numbers.<\/li>\r\n \t<li>If there is a denominator in the function\u2019s equation, exclude values in the domain that force the denominator to be zero.<\/li>\r\n \t<li>If there is an even root, consider excluding values that would make the quantity under the radical negative.<\/li>\r\n<\/ol>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Domain of a Function<\/h3>\r\nFind the domain of the function [latex]f\\left(x\\right)={x}^{2}-1[\/latex].\r\n\r\n[reveal-answer q=\"100687\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"100687\"]\r\n\r\nThe input value, shown by the variable [latex]x[\/latex] in the equation, is squared and then the result is lowered by one. Any real number may be squared and then be lowered by one, so there are no restrictions on the domain of this function. The domain is the set of real numbers.\r\n\r\nIn interval form the domain of [latex]f[\/latex] is [latex]\\left(-\\infty ,\\infty \\right)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Domain of a Rational Function<\/h3>\r\nFind the domain of the function [latex]f\\left(x\\right)=\\dfrac{x+1}{2-x}[\/latex].\r\n\r\n[reveal-answer q=\"759017\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"759017\"]\r\n\r\nWhen there is a denominator, we want to include only values of the input that do not force the denominator to be zero. So, we will set the denominator equal to 0 and solve for [latex]x[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}2-x&amp;=0 \\\\ -x&amp;=-2 \\\\ x&amp;=2 \\end{align}[\/latex]<\/p>\r\nNow, we will exclude 2 from the domain. The answers are all real numbers where [latex]x&lt;2[\/latex] or [latex]x&gt;2[\/latex]. We can use a symbol known as the union, [latex]\\cup [\/latex], to combine the two sets. In interval notation, we write the solution: [latex]\\left(\\mathrm{-\\infty },2\\right)\\cup \\left(2,\\infty \\right)[\/latex].\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18193532\/CNX_Precalc_Figure_01_02_028n2.jpg\" alt=\"Line graph of x=!2.\" width=\"487\" height=\"164\" \/>\r\n\r\nIn interval form, the domain of [latex]f[\/latex] is [latex]\\left(-\\infty ,2\\right)\\cup \\left(2,\\infty \\right)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWatch the following video to see more examples of how to find the domain of a rational function.\r\n\r\n<iframe src=\"\/\/plugin.3playmedia.com\/show?mf=6454925&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=v0IhvIzCc_I&amp;video_target=tpm-plugin-j4jfaxx4-v0IhvIzCc_I\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe>\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/ExTheDomainOfRationalFunctions_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for \"Ex: The Domain of Rational Functions\" here (opens in new window)<\/a>.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFind the domain of the function: [latex]f\\left(x\\right)=\\dfrac{1+4x}{2x - 1}[\/latex].\r\n\r\n[reveal-answer q=\"307426\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"307426\"]\r\n\r\n[latex]\\left(-\\infty ,\\frac{1}{2}\\right)\\cup \\left(\\frac{1}{2},\\infty \\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Domain of a Function with an Even Root<\/h3>\r\nFind the domain of the function [latex]f\\left(x\\right)=\\sqrt{7-x}[\/latex].\r\n\r\n[reveal-answer q=\"722013\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"722013\"]\r\n\r\nWhen there is an even root in the formula, we exclude any real numbers that result in a negative number in the radicand.\r\n\r\nSet the radicand greater than or equal to zero and solve for [latex]x[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{align}7-x&amp;\\ge 0 \\\\ -x&amp;\\ge -7 \\\\ x&amp;\\le 7 \\end{align}[\/latex]<\/p>\r\nNow, we will exclude any number greater than 7 from the domain. The answers are all real numbers less than or equal to [latex]7[\/latex], or [latex]\\left(-\\infty ,7\\right][\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe next video gives more examples of how to define the domain of a function that contains an even root.\r\n\r\n<iframe src=\"\/\/plugin.3playmedia.com\/show?mf=6454926&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=lj_JB8sfyIM&amp;video_target=tpm-plugin-umc0ulso-lj_JB8sfyIM\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe>\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/ExDomainAndRangeOfSquareRootFunctions_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for \"Ex: Domain and Range of Square Root Functions\" here (opens in new window)<\/a>.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFind the domain of the function [latex]f\\left(x\\right)=\\sqrt{5+2x}[\/latex].\r\n\r\n[reveal-answer q=\"643325\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"643325\"][latex]\\left[-\\frac{5}{2},\\infty \\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Solve Quadratic Equations by Factoring<\/h2>\r\n<strong><em>(also in Module 3,\u00a0Skills Review for Differentiation Rules)<\/em><\/strong>\r\n\r\nOften, the easiest method of solving a quadratic equation is by\u00a0<strong>factoring<\/strong>. When solving a quadratic equation by factoring, you must first set the equation equal to zero, factor, and then set each of the factors equal to 0 and solve for the variable.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving Quadratic Equations by Factoring<\/h3>\r\nFactor and solve the equation: [latex]{x}^{2}+x - 6=0[\/latex].\r\n\r\n[reveal-answer q=\"16111\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"16111\"]\r\n\r\nThe equation is already set equal to zero, so we factor. The factored form of the equation is:\r\n<p style=\"text-align: center;\">[latex]\\left(x - 2\\right)\\left(x+3\\right)=0[\/latex]<\/p>\r\nNow, set each factor equal to zero and solve.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\left(x - 2\\right)\\left(x+3\\right)=0\\hfill \\\\ \\left(x - 2\\right)=0\\hfill \\\\ x=2\\hfill \\\\ \\left(x+3\\right)=0\\hfill \\\\ x=-3\\hfill \\end{array}[\/latex]<\/p>\r\nThe two solutions are [latex]x=2[\/latex] and [latex]x=-3[\/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]2029[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Solve Rational Equations Set Equal to Zero<\/h2>\r\nEquations that contain rational expressions are called <strong>rational equations<\/strong>. For example, [latex] \\frac{2x+1}{4}=\\frac{x}{3}[\/latex] is a rational equation.\u00a0Specifically, for the sake of calculus, you will often take derivatives that are rational expressions and set them equal to 0.\r\n\r\nAs a bit of a hint, almost always, the best way to determine where a rational expression (fraction) is equal to 0 is to set the numerator of the fraction equal to 0.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example: Solving A Rational Equation Set Equal to Zero<\/h3>\r\nSolve [latex]\\dfrac{3x+2}{x^2}=0[\/latex].\r\n\r\n[reveal-answer q=\"843520\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"843520\"]\r\n\r\nNote that x cannot equal 0 since this would cause the denominator to equal 0. Now, to solve the equation, set the numerator equal to 0. So, we solve:\r\n<p style=\"text-align: center;\">[latex]3x+2=0[\/latex]<\/p>\r\nSubtract 2 from both sides of the equation to get the term with the variable by itself.\r\n<p style=\"text-align: center;\">[latex]3x=-2[\/latex]<\/p>\r\nDivide both sides by 3 to isolate the variable.\r\n<p style=\"text-align: center;\">[latex]x=-\\dfrac{2}{3}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve [latex]\\dfrac{9x+11}{2x}=0[\/latex].\r\n\r\n[reveal-answer q=\"455516\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"455516\"]\r\n\r\n[latex]x=-\\dfrac{11}{9}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Solve Radical Equations<\/h2>\r\nAn equation that contains a <strong>radical expression<\/strong> is called a <strong>radical equation<\/strong>. Solving radical equations requires applying the rules of exponents and following some basic algebraic principles. In some cases, it also requires looking out for errors generated by raising unknown quantities to an even power.\r\n<div class=\"textbox shaded\">\r\n<h3>Solving Radical Equations<\/h3>\r\nFollow the following four steps to solve radical equations.\r\n<ol>\r\n \t<li>Isolate the radical expression.<\/li>\r\n \t<li>Square both sides of the equation: If [latex]x=y[\/latex], then [latex]x^{2}=y^{2}[\/latex].<\/li>\r\n \t<li>Once the radical is removed, solve for the unknown.<\/li>\r\n \t<li>Check all answers.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example: Solving Radical Equations<\/h3>\r\nSolve. [latex] \\sqrt{x}-3=5[\/latex]\r\n\r\n[reveal-answer q=\"946356\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"946356\"]\r\n\r\nAdd\u00a0[latex]3[\/latex] to both sides to isolate the variable term on the left side of the equation.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{r}\\sqrt{x}-3\\,\\,\\,=\\,\\,\\,5\\\\\\underline{+3\\,\\,\\,\\,\\,\\,\\,+3}\\end{array}[\/latex]<\/p>\r\nCombine like terms.\r\n<p style=\"text-align: center;\">[latex] \\sqrt{x}=8[\/latex]<\/p>\r\nSquare both sides to remove the radical since [latex] {{(\\sqrt{x})}^{2}}=x[\/latex]. Make sure to square the\u00a0[latex]8[\/latex] also! Then simplify.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}{{(\\sqrt{x})}^{2}}={{8}^{2}}\\\\x=64\\end{array}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nTo check your solution, you can substitute\u00a0[latex]64[\/latex] in for <i>x<\/i> in the original equation. Does [latex] \\sqrt{64}-3=5[\/latex]? Yes\u2014the square root of\u00a0[latex]64[\/latex] is\u00a0[latex]8[\/latex], and [latex]8\u22123=5[\/latex].\r\n\r\nNotice how you combined like terms and then squared both <i>sides<\/i> of the equation in this problem. This is a standard method for removing a radical from an equation. It is important to isolate a radical on one side of the equation and simplify as much as possible <i>before<\/i> squaring. The fewer terms there are before squaring, the fewer additional terms will be generated by the process of squaring.\r\n\r\nIn the example above, only the variable <i>x<\/i> was underneath the radical. Sometimes you will need to solve an equation that contains multiple terms underneath a radical. Follow the same steps to solve these, but pay attention to a critical point\u2014square both <i>sides<\/i> of an equation, not individual <i>terms<\/i>.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example: Solving Radical Equations<\/h3>\r\nSolve. [latex] 1+\\sqrt{2x+3}=6[\/latex]\r\n\r\n[reveal-answer q=\"479262\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"479262\"]\r\n\r\nBegin by subtracting\u00a0[latex]1[\/latex] from both sides in order to isolate the radical term. Then square both sides to remove the binomial from the radical.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{r}1+\\sqrt{2x+3}-1=6-1\\\\\\sqrt{2x+3}=5\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\{{\\left( \\sqrt{2x+3} \\right)}^{2}}={{\\left( 5 \\right)}^{2}}\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nSimplify the equation and solve for <i>x<\/i>.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{r}2x+3=25\\\\2x=22\\\\x=11\\end{array}[\/latex]<\/p>\r\nCheck your answer. Substituting\u00a0[latex]11[\/latex] for <i>x<\/i> in the original equation yields a true statement, so the solution is correct.\r\n<p style=\"text-align: center;\">[latex] \\begin{array}{r}1+\\sqrt{2(11)+3}=6\\\\1+\\sqrt{22+3}=6\\\\1+\\sqrt{25}=6\\\\1+5=6\\\\6=6\\end{array}[\/latex]<\/p>\r\n[latex] x=11[\/latex] is the solution for [latex] 1+\\sqrt{2x+3}=6[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video, you will see two more examples that are similar to the ones above.\r\n\r\n<iframe src=\"\/\/plugin.3playmedia.com\/show?mf=6454927&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=tT0Zwsto6AQ&amp;video_target=tpm-plugin-sm4lx3pt-tT0Zwsto6AQ\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe>\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/Ex1SolveABasicRadicalEquationSquareRoots_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for \"Ex 1: Solve a Basic Radical Equation - Square Roots\" here (opens in new window)<\/a>.\r\n<h2>Solve Trigonometric Equations<\/h2>\r\n<strong>Trigonometric equations<\/strong> are, as the name implies, equations that involve trigonometric functions. We will often solve a trigonometric equation over a specified interval. However, we will sometimes 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. 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.\r\n\r\nOne quick way to solve trigonometric equations is to use the unit circle. Remember that the unit circle tells us the value of cosine and sine at any of the given angle measures seen below. The first coordinate in each ordered pair is the value of cosine at the given angle measure, while the second coordinate in each ordered pair is the value of sine at the given angle measure.\r\n\r\n<img class=\"aligncenter wp-image-12625 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003609\/f-d-43392176e093fa07f39e1f3687226d4d751b809be928c16abac5dcb3-IMAGE-IMAGE.png\" alt=\"f-d-43392176e093fa07f39e1f3687226d4d751b809be928c16abac5dcb3+IMAGE+IMAGE\" width=\"800\" height=\"728\" \/>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving A Trigonometric Equation Containing Cosine<\/h3>\r\nFind the exact solutions of 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\nNote: Most of the time, your interval will be restricted to\u00a0 [latex]\\left[0,2\\pi \\right)[\/latex] where you would then say your only solutions are\u00a0[latex] \\theta=\\frac{\\pi }{3}[\/latex] and [latex] \\theta=\\frac{5\\pi }{3}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving A Trigonometric Equation Containing sine<\/h3>\r\nFind the exact solutions of the equation [latex]\\sin \\theta=\\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 [latex] \\theta[\/latex] 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] \\theta=\\frac{\\pi }{6}[\/latex] and [latex] \\theta=\\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] \\theta=\\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\nNote: Most of the time, your interval will be restricted to\u00a0 [latex]\\left[0,2\\pi \\right)[\/latex] where you would then say your only solutions are\u00a0[latex] \\theta=\\frac{\\pi }{6}[\/latex] and [latex] \\theta=\\frac{5\\pi }{6}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving A Trigonometric Equation Containing Cosine<\/h3>\r\nFind the exact solutions of the equation [latex]2\\cos x -3=-5[\/latex] on the interval\u00a0[latex]\\left[0,2\\pi \\right)[\/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\nFind the exact solutions of the 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>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li><span data-sheets-value=\"{&quot;1&quot;:2,&quot;2&quot;:&quot;Find the domain of a function algebraically&quot;}\" data-sheets-userformat=\"{&quot;2&quot;:4609,&quot;3&quot;:{&quot;1&quot;:0},&quot;12&quot;:0,&quot;15&quot;:&quot;Work Sans&quot;}\">Find the domain of a function algebraically<\/span><\/li>\n<li>Use the zero product principle to solve quadratic equations that can be factored<\/li>\n<li>Solve rational equations set equal to zero<\/li>\n<li>Solve for a variable in a square root equation<\/li>\n<li>Solve trigonometric equations<\/li>\n<\/ul>\n<\/div>\n<p>In the Maxima and Minima and the Mean Value Theorem sections, we will find the domain of functions and their derivatives. Derivatives will also be set equal to 0 or some specific value in both sections which will lead to having to solve various types of equations. Here we will review how to find the domain of functions and how to solve various types of equations.<\/p>\n<h2>Find the Domain of a Function<\/h2>\n<p>Recall that the domain of a function is all input values where the function is defined. When given a function equation, you must remember the following three cases:<\/p>\n<ol>\n<li>If the function has no denominator or an even root, in most cases, the domain will be all real numbers.<\/li>\n<li>If there is a denominator in the function\u2019s equation, exclude values in the domain that force the denominator to be zero.<\/li>\n<li>If there is an even root, consider excluding values that would make the quantity under the radical negative.<\/li>\n<\/ol>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Domain of a Function<\/h3>\n<p>Find the domain of the function [latex]f\\left(x\\right)={x}^{2}-1[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q100687\">Show Solution<\/span><\/p>\n<div id=\"q100687\" class=\"hidden-answer\" style=\"display: none\">\n<p>The input value, shown by the variable [latex]x[\/latex] in the equation, is squared and then the result is lowered by one. Any real number may be squared and then be lowered by one, so there are no restrictions on the domain of this function. The domain is the set of real numbers.<\/p>\n<p>In interval form the domain of [latex]f[\/latex] is [latex]\\left(-\\infty ,\\infty \\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Domain of a Rational Function<\/h3>\n<p>Find the domain of the function [latex]f\\left(x\\right)=\\dfrac{x+1}{2-x}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q759017\">Show Solution<\/span><\/p>\n<div id=\"q759017\" class=\"hidden-answer\" style=\"display: none\">\n<p>When there is a denominator, we want to include only values of the input that do not force the denominator to be zero. So, we will set the denominator equal to 0 and solve for [latex]x[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}2-x&=0 \\\\ -x&=-2 \\\\ x&=2 \\end{align}[\/latex]<\/p>\n<p>Now, we will exclude 2 from the domain. The answers are all real numbers where [latex]x<2[\/latex] or [latex]x>2[\/latex]. We can use a symbol known as the union, [latex]\\cup[\/latex], to combine the two sets. In interval notation, we write the solution: [latex]\\left(\\mathrm{-\\infty },2\\right)\\cup \\left(2,\\infty \\right)[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18193532\/CNX_Precalc_Figure_01_02_028n2.jpg\" alt=\"Line graph of x=!2.\" width=\"487\" height=\"164\" \/><\/p>\n<p>In interval form, the domain of [latex]f[\/latex] is [latex]\\left(-\\infty ,2\\right)\\cup \\left(2,\\infty \\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see more examples of how to find the domain of a rational function.<\/p>\n<p><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=6454925&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=v0IhvIzCc_I&amp;video_target=tpm-plugin-j4jfaxx4-v0IhvIzCc_I\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/ExTheDomainOfRationalFunctions_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for &#8220;Ex: The Domain of Rational Functions&#8221; here (opens in new window)<\/a>.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Find the domain of the function: [latex]f\\left(x\\right)=\\dfrac{1+4x}{2x - 1}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q307426\">Show Solution<\/span><\/p>\n<div id=\"q307426\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\left(-\\infty ,\\frac{1}{2}\\right)\\cup \\left(\\frac{1}{2},\\infty \\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Domain of a Function with an Even Root<\/h3>\n<p>Find the domain of the function [latex]f\\left(x\\right)=\\sqrt{7-x}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q722013\">Show Solution<\/span><\/p>\n<div id=\"q722013\" class=\"hidden-answer\" style=\"display: none\">\n<p>When there is an even root in the formula, we exclude any real numbers that result in a negative number in the radicand.<\/p>\n<p>Set the radicand greater than or equal to zero and solve for [latex]x[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}7-x&\\ge 0 \\\\ -x&\\ge -7 \\\\ x&\\le 7 \\end{align}[\/latex]<\/p>\n<p>Now, we will exclude any number greater than 7 from the domain. The answers are all real numbers less than or equal to [latex]7[\/latex], or [latex]\\left(-\\infty ,7\\right][\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The next video gives more examples of how to define the domain of a function that contains an even root.<\/p>\n<p><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=6454926&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=lj_JB8sfyIM&amp;video_target=tpm-plugin-umc0ulso-lj_JB8sfyIM\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/ExDomainAndRangeOfSquareRootFunctions_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for &#8220;Ex: Domain and Range of Square Root Functions&#8221; here (opens in new window)<\/a>.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Find the domain of the function [latex]f\\left(x\\right)=\\sqrt{5+2x}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q643325\">Show Solution<\/span><\/p>\n<div id=\"q643325\" class=\"hidden-answer\" style=\"display: none\">[latex]\\left[-\\frac{5}{2},\\infty \\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Solve Quadratic Equations by Factoring<\/h2>\n<p><strong><em>(also in Module 3,\u00a0Skills Review for Differentiation Rules)<\/em><\/strong><\/p>\n<p>Often, the easiest method of solving a quadratic equation is by\u00a0<strong>factoring<\/strong>. When solving a quadratic equation by factoring, you must first set the equation equal to zero, factor, and then set each of the factors equal to 0 and solve for the variable.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Solving Quadratic Equations by Factoring<\/h3>\n<p>Factor and solve the equation: [latex]{x}^{2}+x - 6=0[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q16111\">Show Solution<\/span><\/p>\n<div id=\"q16111\" class=\"hidden-answer\" style=\"display: none\">\n<p>The equation is already set equal to zero, so we factor. The factored form of the equation is:<\/p>\n<p style=\"text-align: center;\">[latex]\\left(x - 2\\right)\\left(x+3\\right)=0[\/latex]<\/p>\n<p>Now, set each factor equal to zero and solve.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\left(x - 2\\right)\\left(x+3\\right)=0\\hfill \\\\ \\left(x - 2\\right)=0\\hfill \\\\ x=2\\hfill \\\\ \\left(x+3\\right)=0\\hfill \\\\ x=-3\\hfill \\end{array}[\/latex]<\/p>\n<p>The two solutions are [latex]x=2[\/latex] and [latex]x=-3[\/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=\"ohm2029\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=2029&theme=oea&iframe_resize_id=ohm2029&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Solve Rational Equations Set Equal to Zero<\/h2>\n<p>Equations that contain rational expressions are called <strong>rational equations<\/strong>. For example, [latex]\\frac{2x+1}{4}=\\frac{x}{3}[\/latex] is a rational equation.\u00a0Specifically, for the sake of calculus, you will often take derivatives that are rational expressions and set them equal to 0.<\/p>\n<p>As a bit of a hint, almost always, the best way to determine where a rational expression (fraction) is equal to 0 is to set the numerator of the fraction equal to 0.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example: Solving A Rational Equation Set Equal to Zero<\/h3>\n<p>Solve [latex]\\dfrac{3x+2}{x^2}=0[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q843520\">Show Solution<\/span><\/p>\n<div id=\"q843520\" class=\"hidden-answer\" style=\"display: none\">\n<p>Note that x cannot equal 0 since this would cause the denominator to equal 0. Now, to solve the equation, set the numerator equal to 0. So, we solve:<\/p>\n<p style=\"text-align: center;\">[latex]3x+2=0[\/latex]<\/p>\n<p>Subtract 2 from both sides of the equation to get the term with the variable by itself.<\/p>\n<p style=\"text-align: center;\">[latex]3x=-2[\/latex]<\/p>\n<p>Divide both sides by 3 to isolate the variable.<\/p>\n<p style=\"text-align: center;\">[latex]x=-\\dfrac{2}{3}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve [latex]\\dfrac{9x+11}{2x}=0[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q455516\">Show Solution<\/span><\/p>\n<div id=\"q455516\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x=-\\dfrac{11}{9}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Solve Radical Equations<\/h2>\n<p>An equation that contains a <strong>radical expression<\/strong> is called a <strong>radical equation<\/strong>. Solving radical equations requires applying the rules of exponents and following some basic algebraic principles. In some cases, it also requires looking out for errors generated by raising unknown quantities to an even power.<\/p>\n<div class=\"textbox shaded\">\n<h3>Solving Radical Equations<\/h3>\n<p>Follow the following four steps to solve radical equations.<\/p>\n<ol>\n<li>Isolate the radical expression.<\/li>\n<li>Square both sides of the equation: If [latex]x=y[\/latex], then [latex]x^{2}=y^{2}[\/latex].<\/li>\n<li>Once the radical is removed, solve for the unknown.<\/li>\n<li>Check all answers.<\/li>\n<\/ol>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Example: Solving Radical Equations<\/h3>\n<p>Solve. [latex]\\sqrt{x}-3=5[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q946356\">Show Solution<\/span><\/p>\n<div id=\"q946356\" class=\"hidden-answer\" style=\"display: none\">\n<p>Add\u00a0[latex]3[\/latex] to both sides to isolate the variable term on the left side of the equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}\\sqrt{x}-3\\,\\,\\,=\\,\\,\\,5\\\\\\underline{+3\\,\\,\\,\\,\\,\\,\\,+3}\\end{array}[\/latex]<\/p>\n<p>Combine like terms.<\/p>\n<p style=\"text-align: center;\">[latex]\\sqrt{x}=8[\/latex]<\/p>\n<p>Square both sides to remove the radical since [latex]{{(\\sqrt{x})}^{2}}=x[\/latex]. Make sure to square the\u00a0[latex]8[\/latex] also! Then simplify.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}{{(\\sqrt{x})}^{2}}={{8}^{2}}\\\\x=64\\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>To check your solution, you can substitute\u00a0[latex]64[\/latex] in for <i>x<\/i> in the original equation. Does [latex]\\sqrt{64}-3=5[\/latex]? Yes\u2014the square root of\u00a0[latex]64[\/latex] is\u00a0[latex]8[\/latex], and [latex]8\u22123=5[\/latex].<\/p>\n<p>Notice how you combined like terms and then squared both <i>sides<\/i> of the equation in this problem. This is a standard method for removing a radical from an equation. It is important to isolate a radical on one side of the equation and simplify as much as possible <i>before<\/i> squaring. The fewer terms there are before squaring, the fewer additional terms will be generated by the process of squaring.<\/p>\n<p>In the example above, only the variable <i>x<\/i> was underneath the radical. Sometimes you will need to solve an equation that contains multiple terms underneath a radical. Follow the same steps to solve these, but pay attention to a critical point\u2014square both <i>sides<\/i> of an equation, not individual <i>terms<\/i>.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example: Solving Radical Equations<\/h3>\n<p>Solve. [latex]1+\\sqrt{2x+3}=6[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q479262\">Show Solution<\/span><\/p>\n<div id=\"q479262\" class=\"hidden-answer\" style=\"display: none\">\n<p>Begin by subtracting\u00a0[latex]1[\/latex] from both sides in order to isolate the radical term. Then square both sides to remove the binomial from the radical.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}1+\\sqrt{2x+3}-1=6-1\\\\\\sqrt{2x+3}=5\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\{{\\left( \\sqrt{2x+3} \\right)}^{2}}={{\\left( 5 \\right)}^{2}}\\,\\,\\,\\end{array}[\/latex]<\/p>\n<p>Simplify the equation and solve for <i>x<\/i>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}2x+3=25\\\\2x=22\\\\x=11\\end{array}[\/latex]<\/p>\n<p>Check your answer. Substituting\u00a0[latex]11[\/latex] for <i>x<\/i> in the original equation yields a true statement, so the solution is correct.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{r}1+\\sqrt{2(11)+3}=6\\\\1+\\sqrt{22+3}=6\\\\1+\\sqrt{25}=6\\\\1+5=6\\\\6=6\\end{array}[\/latex]<\/p>\n<p>[latex]x=11[\/latex] is the solution for [latex]1+\\sqrt{2x+3}=6[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video, you will see two more examples that are similar to the ones above.<\/p>\n<p><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=6454927&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=tT0Zwsto6AQ&amp;video_target=tpm-plugin-sm4lx3pt-tT0Zwsto6AQ\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/Ex1SolveABasicRadicalEquationSquareRoots_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for &#8220;Ex 1: Solve a Basic Radical Equation &#8211; Square Roots&#8221; here (opens in new window)<\/a>.<\/p>\n<h2>Solve Trigonometric Equations<\/h2>\n<p><strong>Trigonometric equations<\/strong> are, as the name implies, equations that involve trigonometric functions. We will often solve a trigonometric equation over a specified interval. However, we will sometimes 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. 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.<\/p>\n<p>One quick way to solve trigonometric equations is to use the unit circle. Remember that the unit circle tells us the value of cosine and sine at any of the given angle measures seen below. The first coordinate in each ordered pair is the value of cosine at the given angle measure, while the second coordinate in each ordered pair is the value of sine at the given angle measure.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-12625 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/3675\/2018\/09\/27003609\/f-d-43392176e093fa07f39e1f3687226d4d751b809be928c16abac5dcb3-IMAGE-IMAGE.png\" alt=\"f-d-43392176e093fa07f39e1f3687226d4d751b809be928c16abac5dcb3+IMAGE+IMAGE\" width=\"800\" height=\"728\" \/><\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Solving A Trigonometric Equation Containing Cosine<\/h3>\n<p>Find the exact solutions of 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<p>Note: Most of the time, your interval will be restricted to\u00a0 [latex]\\left[0,2\\pi \\right)[\/latex] where you would then say your only solutions are\u00a0[latex]\\theta=\\frac{\\pi }{3}[\/latex] and [latex]\\theta=\\frac{5\\pi }{3}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving A Trigonometric Equation Containing sine<\/h3>\n<p>Find the exact solutions of the equation [latex]\\sin \\theta=\\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 [latex]\\theta[\/latex] 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]\\theta=\\frac{\\pi }{6}[\/latex] and [latex]\\theta=\\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]\\theta=\\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<p>Note: Most of the time, your interval will be restricted to\u00a0 [latex]\\left[0,2\\pi \\right)[\/latex] where you would then say your only solutions are\u00a0[latex]\\theta=\\frac{\\pi }{6}[\/latex] and [latex]\\theta=\\frac{5\\pi }{6}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving A Trigonometric Equation Containing Cosine<\/h3>\n<p>Find the exact solutions of the equation [latex]2\\cos x -3=-5[\/latex] on the interval\u00a0[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=\"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>Find the exact solutions of the 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\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-3894\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Modification and Revision . <strong>Provided by<\/strong>: Lumen Learning. <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 class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>College Algebra Corequisite. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/\">https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Precalculus. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/precalculus\/\">https:\/\/courses.lumenlearning.com\/precalculus\/<\/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":17533,"menu_order":3,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Modification and Revision \",\"author\":\"\",\"organization\":\"Lumen 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