{"id":2012,"date":"2025-07-31T23:43:56","date_gmt":"2025-07-31T23:43:56","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=2012"},"modified":"2025-10-16T17:57:45","modified_gmt":"2025-10-16T17:57:45","slug":"solving-trigonometric-equations-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/solving-trigonometric-equations-learn-it-3\/","title":{"raw":"Solving Trigonometric Equations: Learn It 3","rendered":"Solving Trigonometric Equations: Learn It 3"},"content":{"raw":"

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\r\n
Solve the problem exactly: [latex]2{\\sin }^{2}\\theta -1=0,0\\le \\theta <2\\pi [\/latex].[reveal-answer q=\"467313\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"467313\"]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.\r\n

[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<\/section>

Solve the following equation exactly: [latex]\\csc \\theta =-2,0\\le \\theta <4\\pi [\/latex].[reveal-answer q=\"605306\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"605306\"]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].\r\n

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

Solve the equation exactly: [latex]\\tan \\left(\\theta -\\frac{\\pi }{2}\\right)=1,0\\le \\theta <2\\pi [\/latex].[reveal-answer q=\"484899\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"484899\"]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\r\n

[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

[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<\/section>\r\n

Find all solutions for [latex]\\tan x=\\sqrt{3}[\/latex].[reveal-answer q=\"629684\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"629684\"][latex]\\frac{\\pi }{3}\\pm \\pi k[\/latex][\/hidden-answer]<\/section>
[ohm_question hide_question_numbers=1]173739[\/ohm_question]<\/section>
Identify all exact solutions to the equation [latex]2\\left(\\tan x+3\\right)=5+\\tan x,0\\le x<2\\pi [\/latex].[reveal-answer q=\"949694\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"949694\"]We can solve this equation using only algebra. Isolate the expression [latex]\\tan x[\/latex] on the left side of the equals sign.\r\n

[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<\/section><\/div>","rendered":"

Solving Equations Involving a Single Trigonometric Function<\/h2>\n

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

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