{"id":2013,"date":"2025-07-31T23:43:58","date_gmt":"2025-07-31T23:43:58","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=2013"},"modified":"2025-10-16T17:58:40","modified_gmt":"2025-10-16T17:58:40","slug":"solving-trigonometric-equations-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/solving-trigonometric-equations-learn-it-4\/","title":{"raw":"Solving Trigonometric Equations: Learn It 4","rendered":"Solving Trigonometric Equations: Learn It 4"},"content":{"raw":"

Solving Trigonometric Equations in Quadratic Form<\/h2>\r\nSolving a 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\r\n
Solve the equation exactly: [latex]{\\cos }^{2}\\theta +3\\cos \\theta -1=0,0\\le \\theta <2\\pi[\/latex].[reveal-answer q=\"563509\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"563509\"]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\r\n

[latex]{x}^{2}+3x - 1=0[\/latex]<\/p>\r\nThe equation cannot be factored, so we will use the quadratic formula<\/strong> [latex]x=\\frac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a}[\/latex].\r\n

[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>\r\nReplace [latex]x[\/latex] with [latex]\\cos \\theta[\/latex], and solve. Thus,\r\n

[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

[latex]\\begin{align}\\theta &={\\cos }^{-1}\\left(\\frac{-3+\\sqrt{13}}{2}\\right) \\\\ &\\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

[latex]\\begin{align}\\theta &=2\\pi -{\\cos }^{-1}\\left(\\frac{-3+\\sqrt{13}}{2}\\right) \\\\ &\\approx 5.02 \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>

Solve the equation exactly: [latex]2{\\sin }^{2}\\theta -5\\sin \\theta +3=0,0\\le \\theta \\le 2\\pi[\/latex].[reveal-answer q=\"693370\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"693370\"]Using grouping, this quadratic can be factored. Either make the real substitution, [latex]\\sin \\theta =u[\/latex], or imagine it, as we factor:\r\n

[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

[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\r\nAnalysis of the Solution<\/strong>\r\n\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<\/section>\r\n

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.][reveal-answer q=\"608693\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"608693\"][latex]\\cos \\theta =-1,\\theta =\\pi [\/latex][\/hidden-answer]<\/section>
[ohm_question hide_question_numbers=1]149010[\/ohm_question]<\/section>
Solve exactly:\r\n

[latex]2{\\sin }^{2}\\theta +\\sin \\theta =0;0\\le \\theta <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

[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

[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

[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 <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

[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\nAnalysis of the Solution<\/strong>\r\n\r\nWe can see the solutions on the graph in Figure 3. On the interval [latex]0\\le \\theta <2\\pi[\/latex], the graph crosses the 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\"]\"Graph Figure 3<\/b>[\/caption]\r\n\r\nWe can verify the solutions on the unit circle<\/a> in Sum and Difference Identities\u00a0as well.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

Solve the equation quadratic in form exactly: [latex]2{\\sin }^{2}\\theta -3\\sin \\theta +1=0,0\\le \\theta <2\\pi[\/latex].[reveal-answer q=\"699307\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"699307\"]We can factor using grouping. Solution values of [latex]\\theta[\/latex] can be found on the unit circle:\r\n

[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<\/section>

Solve the quadratic equation [latex]2{\\cos }^{2}\\theta +\\cos \\theta =0[\/latex].[reveal-answer q=\"736189\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"736189\"][latex]\\frac{\\pi }{2},\\frac{2\\pi }{3},\\frac{4\\pi }{3},\\frac{3\\pi }{2}[\/latex][\/hidden-answer]<\/section><\/div>","rendered":"

Solving Trigonometric Equations in Quadratic Form<\/h2>\n

Solving a 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

Solve the equation exactly: [latex]{\\cos }^{2}\\theta +3\\cos \\theta -1=0,0\\le \\theta <2\\pi[\/latex].\n\n