{"id":2015,"date":"2025-07-31T23:44:01","date_gmt":"2025-07-31T23:44:01","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=2015"},"modified":"2025-10-16T17:58:52","modified_gmt":"2025-10-16T17:58:52","slug":"solving-trigonometric-equations-learn-it-5","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/solving-trigonometric-equations-learn-it-5\/","title":{"raw":"Solving Trigonometric Equations: Learn It 5","rendered":"Solving Trigonometric Equations: Learn It 5"},"content":{"raw":"

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\r\n
Use identities to solve exactly the trigonometric equation over the interval [latex]0\\le x<2\\pi [\/latex].\r\n

[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

[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>\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<\/section>

Solve the equation exactly using a double-angle formula: [latex]\\cos \\left(2\\theta \\right)=\\cos \\theta [\/latex].[reveal-answer q=\"631761\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"631761\"]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:\r\n

[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<\/section>

Solve the equation exactly using an identity: [latex]3\\cos \\theta +3=2{\\sin }^{2}\\theta ,0\\le \\theta <2\\pi [\/latex].[reveal-answer q=\"468404\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"468404\"]If we rewrite the right side, we can write the equation in terms of cosine:\r\n

[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>\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<\/section>

[ohm_question hide_question_numbers=1]149012[\/ohm_question]<\/section>\r\n

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 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 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\r\n
Solve exactly: [latex]\\cos \\left(2x\\right)=\\frac{1}{2}[\/latex] on [latex]\\left[0,2\\pi \\right)[\/latex].[reveal-answer q=\"299198\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"299198\"]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.\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

[latex]\\begin{align} 2x&=\\frac{\\pi }{3}+2\\pi \\\\ &=\\frac{\\pi }{3}+\\frac{6\\pi }{3} \\\\ &=\\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

[latex]\\begin{align} 2x&=\\frac{\\pi }{3}+4\\pi \\\\ &=\\frac{\\pi }{3}+\\frac{12\\pi }{3} \\\\ &=\\frac{13\\pi }{3} \\end{align}[\/latex]<\/p>\r\n[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].\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

[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>\r\nso [latex]x=\\frac{11\\pi }{6}[\/latex].\r\n\r\nOne more rotation yields\r\n

[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>\r\n[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].\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<\/section>","rendered":"

Solving Trigonometric Equations Using Fundamental Identities<\/h2>\n

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

Use identities to solve exactly the trigonometric equation over the interval [latex]0\\le x<2\\pi[\/latex].\n\n\n

[latex]\\cos x\\cos \\left(2x\\right)+\\sin x\\sin \\left(2x\\right)=\\frac{\\sqrt{3}}{2}[\/latex]<\/p>\n