Learning Objectives
In this section, you will:
- Solve linear trigonometric equations in sine and cosine.
- Solve equations involving a single trigonometric function.
- Solve trigonometric equations that are quadratic in form.
- Solve trigonometric equations using fundamental identities.
- Solve trigonometric equations with multiple angles.
Let’s Get Started…
Figure 1 Egyptian pyramids standing near a modern city. (credit: Oisin Mulvihill)
Thales of Miletus (circa 625–547 BC) is known as the founder of geometry. The legend is that he calculated the height of the Great Pyramid of Giza in Egypt using the theory of similar triangles, which he developed by measuring the shadow of his staff. Based on proportions, this theory has applications in a number of areas, including fractal geometry, engineering, and architecture. Often, the angle of elevation and the angle of depression are found using similar triangles. In this section, we begin our study of trigonometric equations to study real-world scenarios such as the finding the dimensions of the pyramids.
Solving Linear Trigonometric Equations in Sine and Cosine
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 period of both the sine function and the cosine function is[latex]\,2\pi .[/latex] In other words, every[latex]\,2\pi \,[/latex]units, the [latex]y-[/latex]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 \text{:}[/latex]
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 trigonometric identities.
EXAMPLE 1
Solving a Linear Trigonometric Equation Involving the Cosine Function
Find all possible exact solutions for the equation[latex]\,\mathrm{cos}\left(\theta\right) =\frac{1}{2}.[/latex]
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Solution
From the unit circle, we know that
[latex]\begin{array}{ccc}\hfill \mathrm{cos}\left(\theta\right) & =& \frac{1}{2}\hfill \\ \hfill \theta & =& \frac{\pi }{3},\frac{5\pi }{3}\hfill \end{array}[/latex]
These are the solutions in the interval[latex]\,\left[0,2\pi \right].\,[/latex]All possible solutions are given by
[latex]\theta =\frac{\pi }{3}±2k\pi \; \; \; \text{and} \; \; \; \theta =\frac{5\pi }{3}±2k\pi[/latex]
where[latex]\,k\,[/latex]is an integer.
EXAMPLE 2
Solving a Linear Equation Involving the Sine Function
Find all possible exact solutions for the equation[latex]\,\mathrm{sin}\left(t\right)=\frac{1}{2}.[/latex]
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Solution
Solving for all possible values of [latex]t[/latex] means that solutions include angles beyond the period of[latex]\,2\pi .\,[/latex]From Figure 2, 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
[latex]t=\frac{\pi }{6}±2\pi k \; \; \; \text{and} \; \; \; t=\frac{5\pi }{6}±2\pi k[/latex]
where[latex]\,k\,[/latex]is an integer.
Figure 2
How To
Given a trigonometric equation, solve using algebra.
- Look for a pattern that suggests an algebraic property, such as the difference of squares or a factoring opportunity.
- Substitute the trigonometric expression with a single variable, such as[latex]\,x\,[/latex]or[latex]\,u.[/latex]
- Solve the equation the same way an algebraic equation would be solved.
- Substitute the trigonometric expression back in for the variable in the resulting expressions.
- Solve for the angle.
EXAMPLE 3
Solve the Linear Trigonometric Equation
Solve the equation exactly:[latex]\,2\,\mathrm{cos}\left(\theta\right) -3=-5,\; \; \; 0\le \theta <2\pi .[/latex]
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Solution
Use algebraic techniques to solve the equation.
[latex]\begin{array}{ccc}\hfill 2\,\mathrm{cos}\left(\theta\right) -3& =& -5\hfill \\ \hfill 2\,\mathrm{cos}\left(\theta\right) & =& -2\hfill \\ \hfill \mathrm{cos}\left(\theta\right) & =& -1\hfill \\ \hfill \theta & =& \pi \hfill \end{array}[/latex]
Try It #1
Solve exactly the following linear equation on the interval[latex]\,\left[0,2\pi \right):\; \; 2\,\mathrm{sin}\left(x\right)+1=0.[/latex]
Solving Equations Involving a Single Trigonometric Function
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.
EXAMPLE 4
Solving a Problem Involving a Single Trigonometric Function
Solve the problem exactly:[latex]\,2\,{\mathrm{sin}}^{2}\left(\theta\right) -1=0, \; \; \; 0\le \theta <2\pi .[/latex]
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Solution
As this problem is not easily factored, we will solve using the square root property. First, we use algebra to isolate[latex]\,\mathrm{sin}\left(\theta\right) .\,[/latex]Then we will find the angles.
[latex]\begin{array}{ccc}\hfill 2\,{\mathrm{sin}}^{2}\left(\theta\right) -1& =& 0\hfill \\ \hfill \text{ }2\,{\mathrm{sin}}^{2}\left(\theta\right) & =& 1\hfill \\ \hfill {\mathrm{sin}}^{2}\left(\theta\right) & =& \frac{1}{2}\hfill \\ \hfill \sqrt{{\mathrm{sin}}^{2}\left(\theta\right) }& =& ±\sqrt{\frac{1}{2}}\hfill \\ \hfill \mathrm{sin}\left(\theta\right) & =& ±\frac{1}{\sqrt{2}}=±\frac{\sqrt{2}}{2}\hfill \\ \hfill \theta & =& \frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}\hfill \end{array}[/latex]
EXAMPLE 5
Solving a Trigonometric Equation Involving Cosecant
Solve the following equation exactly:[latex]\,\mathrm{csc}\left(\theta\right) =-2, \; \; \; 0\le \theta <4\pi .[/latex]
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Solution
Use algebraic techniques to solve the equation.
We want all values of[latex]\,\theta \,[/latex]for which[latex]\,\mathrm{csc}\left(\theta\right) =-2\,[/latex]over the interval[latex]\; 0\le \theta <4\pi .[/latex]
[latex]\begin{array}{ccc}\hfill \mathrm{csc}\left(\theta\right) & =& -2\hfill \\ \hfill \frac{1}{\mathrm{sin}\left(\theta\right) }& =& -2\hfill \\ \hfill \mathrm{sin}\left(\theta\right) & =& -\frac{1}{2}\hfill \\ \hfill \theta & =& \frac{7\pi }{6},\frac{11\pi }{6},\frac{19\pi }{6},\frac{23\pi }{6}\hfill \end{array}[/latex]
Analysis
As[latex]\,\mathrm{sin}\left(\theta\right) =-\frac{1}{2}[/latex], notice that all four solutions are in the third and fourth quadrants.
EXAMPLE 6
Solving an Equation Involving Tangent
Solve the equation exactly:[latex]\,\mathrm{tan}\left(\theta -\frac{\pi }{2}\right)=1, \; \; \; 0\le \theta <2\pi .[/latex]
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Solution
Use algebraic techniques to solve the equation.
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]\,\mathrm{tan}\left(\frac{\pi }{4}\right)=1[/latex], then
[latex]\begin{array}{ccc}\hfill \theta -\frac{\pi }{2}& =& \frac{\pi }{4}\hfill \\ \hfill \theta & =& \frac{3\pi }{4}±k\pi \hfill \end{array}[/latex]
Over the interval [latex]\left[0,2\pi \right)[/latex], we have two solutions:
[latex]\theta =\frac{3\pi }{4}\; \; \; \text{and} \; \; \; \theta =\frac{3\pi }{4}+\pi =\frac{7\pi }{4}[/latex]
Try It #2
Find all solutions for[latex]\,\mathrm{tan}\left(x\right)=\sqrt{3}.[/latex]
EXAMPLE 7
Identify all Solutions to the Equation Involving Tangent
Identify all exact solutions to the equation[latex]\,2\left(\mathrm{tan}\left(x\right)+3\right)=5+\mathrm{tan}\left(x\right), \; \; \; 0\le x\lt 2\pi .[/latex]
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Solution
Use algebraic techniques to solve the equation.
We can solve this equation using only algebra. Isolate the expression[latex]\,\mathrm{tan}\left(x\right)\,[/latex]on the left side of the equals sign
[latex]\begin{array}{ccc}\hfill 2\left(\mathrm{tan}\left(x\right)\right)+2\left(3\right)& =& 5+\mathrm{tan}\left(x\right)\hfill \\ \hfill 2\mathrm{tan}\left(x\right)+6& =& 5+\mathrm{tan}\left(x\right)\hfill \\ \hfill \text{ }2\mathrm{tan}\left(x\right)-\mathrm{tan}\left(x\right)& =& 5-6\hfill \\ \hfill \mathrm{tan}\left(x\right)& =& -1\hfill \end{array}[/latex]
There are two angles on the unit circle that have a tangent value of[latex]\,-1\text{:}\; \; \theta =\frac{3\pi }{4}[/latex] and [latex]\theta =\frac{7\pi }{4}.[/latex]
Solving Trigonometric Equations in Quadratic Form
Solving a quadratic equation 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.
EXAMPLE 8
Solving a Trigonometric Equation in Quadratic Form
Solve the equation exactly:[latex]\,{\mathrm{cos}}^{2}\left(\theta\right) +3\,\mathrm{cos}\left(\theta\right) -1=0, \; \; \; 0\le \theta \lt 2\pi .[/latex]
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Solution
We begin by using substitution and replacing [latex]\mathrm{cos}\left(\theta\right)[/latex] with [latex]x.[/latex] It is not necessary to use substitution, but it may make the problem easier to solve visually. Let [latex]\mathrm{cos}\left(\theta\right) =x.\,[/latex]We have
[latex]{x}^{2}+3x-1=0[/latex]
The equation cannot be factored, so we will use the quadratic formula[latex]\,x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}.[/latex]
[latex]\begin{array}{ccc}\hfill x& =& \frac{-3±\sqrt{{\left(-3\right)}^{2}-4\left(1\right)\left(-1\right)}}{2}\hfill \\ & =& \frac{-3±\sqrt{13}}{2}\hfill \end{array}[/latex]
Replace[latex]\,x\,[/latex]with[latex]\,\mathrm{cos}\left(\theta\right) ,[/latex]and solve.
[latex]\begin{array}{ccc}\hfill \mathrm{cos}\left(\theta\right) & =& \frac{-3±\sqrt{13}}{2}\hfill \\ \hfill \theta & =& {\mathrm{cos}}^{-1}\left(\frac{-3+\sqrt{13}}{2}\right)\hfill \end{array}[/latex]
Note that only the + sign is used. This is because we get an error when we solve [latex]\theta ={\mathrm{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:
[latex]\begin{array}{ccc}\hfill \theta & =& {\mathrm{cos}}^{-1}\left(\frac{-3+\sqrt{13}}{2}\right)\hfill \\ & \approx & 1.26\hfill \end{array}[/latex]
This terminal side of the angle lies in quadrant I. Since cosine is also positive in quadrant IV, the second solution is
[latex]\begin{array}{ccc}\hfill \theta & =& 2\pi -{\mathrm{cos}}^{-1}\left(\frac{-3+\sqrt{13}}{2}\right)\hfill \\ & \approx & \text{ }5.02\hfill \end{array}[/latex]
EXAMPLE 9
Solving a Trigonometric Equation in Quadratic Form by Factoring
Solve the equation exactly:[latex]\,2\,{\mathrm{sin}}^{2}\left(\theta\right) -5\,\mathrm{sin}\left(\theta\right) +3=0, \; \; \; 0\le \theta \le 2\pi .[/latex]
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Solution
Using grouping, this quadratic can be factored. Either make the real substitution,[latex]\,\mathrm{sin}\left(\theta\right) =u,[/latex]or imagine it, as we factor:
[latex]\begin{array}{ccc}\text{ }2\,{\mathrm{sin}}^{2}\left(\theta\right) -5\,\mathrm{sin}\left(\theta\right) +3& =& 0\\ \left(2\,\mathrm{sin}\left(\theta\right) -3\right)\left(\mathrm{sin}\left(\theta\right) -1\right)& =& 0\end{array}[/latex]
Now set each factor equal to zero.
[latex]\begin{array}{ccc}\hfill 2\,\mathrm{sin}\left(\theta\right) -3& =& 0\hfill \\ \hfill 2\,\mathrm{sin}\left(\theta\right) & =& 3\hfill \\ \hfill \mathrm{sin}\left(\theta\right) & =& \frac{3}{2}\hfill \\ & & \\ & & \\ \hfill \mathrm{sin}\left(\theta\right) -1& =& 0\hfill \\ \hfill \mathrm{sin}\left(\theta\right) & =& 1\hfill \end{array}[/latex]
Next solve for [latex]\theta :\mathrm{sin}\left(\theta\right) \ne \frac{3}{2}[/latex], as the range of the sine function is [latex]\left[-1,1\right][/latex]. However, [latex]\mathrm{sin}\left(\theta\right) =1,[/latex] giving the solution [latex]\theta =\frac{\pi }{2}.[/latex]
Analysis
Make sure to check all solutions on the given domain as some factors have no solution.
Try It #3
Solve[latex]\,{\mathrm{sin}}^{2}\left(\theta\right) =2\,\mathrm{cos}\left(\theta\right) +2,\; \; \; 0\le \theta \le 2\pi .[/latex] [Hint: Make a substitution to express the equation only in terms of cosine.]
EXAMPLE 10
Solving a Trigonometric Equation Using Algebra
Solve exactly:
[latex]2\,{\mathrm{sin}}^{2}\left(\theta\right) +\mathrm{sin}\left(\theta\right) =0; \; \; \; 0\le \theta \lt 2\pi[/latex]
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Solution
This problem should appear familiar as it is similar to a quadratic. Let [latex]\mathrm{sin}\left(\theta\right) =x.[/latex] The equation becomes [latex]2{x}^{2}+x=0[/latex]. We begin by factoring:
[latex]\begin{array}{ccc}\hfill 2{x}^{2}+x& =& 0\hfill \\ \hfill x\left(2x+1\right)& =& 0\hfill \end{array}[/latex]
Set each factor equal to zero.
[latex]\begin{array}{ccc}\hfill x& =& 0\hfill \\ \hfill \left(2x+1\right)& =& 0\hfill \\ \hfill x& =& -\frac{1}{2}\hfill \end{array}[/latex]
Then, substitute back into the equation the original expression[latex]\,\mathrm{sin}\left(\theta\right)[/latex] for [latex]x[/latex]. Thus,
[latex]\begin{array}{ccc}\hfill \mathrm{sin}\left(\theta\right) & =& 0\hfill \\ \phantom{\rule{1em}{0ex}}\hfill \theta & =& 0,\pi \hfill \\ \hfill \mathrm{sin}\left(\theta\right) & =& -\frac{1}{2}\hfill \\ \hfill \theta & =& \frac{7\pi }{6},\frac{11\pi }{6}\hfill \end{array}[/latex]
The 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]
If we prefer not to substitute, we can solve the equation by following the same pattern of factoring and setting each factor equal to zero.
[latex]\begin{array}{ccc}\hfill 2\,{\mathrm{sin}}^{2}\left(\theta\right) +\mathrm{sin}\left(\theta\right) & =& 0\hfill \\ \hfill \mathrm{sin}\left(\theta\right) \left(2\mathrm{sin}\left(\theta\right) +1\right)& =& 0\hfill \\ \hfill \mathrm{sin}\left(\theta\right) & =& 0\hfill \\ \hfill \theta & =& 0,\pi \hfill \\ & & \\ \hfill 2\,\mathrm{sin}\left(\theta\right) +1& =& 0\hfill \\ \hfill 2\mathrm{sin}\left(\theta\right) & =& -1\hfill \\ \hfill \mathrm{sin}\left(\theta\right) & =& -\frac{1}{2}\hfill \\ \hfill \theta & =& \frac{7\pi }{6},\frac{11\pi }{6}\hfill \end{array}[/latex]
Analysis
We 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 [latex]x-[/latex]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.
Figure 3
We can verify the solutions on the unit circle in Figure 2 as well.
EXAMPLE 11
Solving a Trigonometric Equation Quadratic in Form
Solve the equation quadratic in form exactly:[latex]\,2\,{\mathrm{sin}}^{2}\left(\theta\right) -3\,\mathrm{sin}\left(\theta\right) +1=0, \; \; \; 0\le \theta \lt 2\pi .[/latex]
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Solution
We can factor using grouping. Solution values of[latex]\,\theta \,[/latex]can be found on the unit circle.
[latex]\begin{array}{ccc}\hfill \left(2\,\mathrm{sin}\left(\theta\right) -1\right)\left(\mathrm{sin}\left(\theta\right) -1\right)& =& 0\hfill \\ \hfill \text{ }2\,\mathrm{sin}\left(\theta\right) -1& =& 0\hfill \\ \hfill \mathrm{sin}\left(\theta\right) & =& \frac{1}{2}\hfill \\ \hfill \theta & =& \frac{\pi }{6},\frac{5\pi }{6}\hfill \\ & & \\ \hfill \mathrm{sin}\,\theta & =& 1\hfill \\ \hfill \theta & =& \frac{\pi }{2}\hfill \end{array}[/latex]
Try It #4
Solve the quadratic equation[latex]\,2\,{\mathrm{cos}}^{2}\left(\theta\right) +\mathrm{cos}\left(\theta\right) =0.[/latex]
Solving Trigonometric Equations Using Fundamental Identities
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.
EXAMPLE 12
Use Identities to Solve an Equation
Use identities to solve exactly the trigonometric equation over the interval[latex]\,0\le x \lt 2\pi .[/latex]
[latex]\mathrm{cos}\left(x\right)\,\mathrm{cos}\left(2x\right)+\mathrm{sin}\left(x\right)\,\mathrm{sin}\left(2x\right)=\frac{\sqrt{3}}{2}[/latex]
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Solution
Notice that the left side of the equation is the difference formula for cosine.
[latex]\begin{array}{cccc}\hfill \mathrm{cos}\left(x\right)\,\mathrm{cos}\left(2x\right)+\mathrm{sin}\left(x\right)\,\mathrm{sin}\left(2x\right)& =& \frac{\sqrt{3}}{2}\hfill & \\ \hfill \mathrm{cos}\left(x-2x\right)& =& \frac{\sqrt{3}}{2}\hfill & \phantom{\rule{2em}{0ex}}\text{Difference formula for cosine}\hfill \\ \hfill \mathrm{cos}\left(-x\right)& =& \frac{\sqrt{3}}{2}\hfill & \phantom{\rule{2em}{0ex}}\text{Use the negative angle identity}.\hfill \\ \hfill \mathrm{cos}\,x& =& \frac{\sqrt{3}}{2}\hfill & \end{array}[/latex]
From the unit circle in Figure 2, we see that[latex]\,\mathrm{cos}\left(x\right)=\frac{\sqrt{3}}{2}[/latex] when [latex]x=\frac{\pi }{6},\frac{11\pi }{6}.[/latex]
EXAMPLE 13
Solving the Equation Using a Double-Angle Formula
Solve the equation exactly using a double-angle formula:[latex]\,\mathrm{cos}\left(2\,\theta \right)=\mathrm{cos}\left(\theta\right).[/latex]
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Solution
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:
[latex]\begin{array}{ccc}\hfill \mathrm{cos}\left(2\theta \right)& =& \mathrm{cos}\left(\theta\right) \hfill \\ \hfill 2{\mathrm{cos}}^{2}\left(\theta\right) -1& =& \mathrm{cos}\left(\theta\right) \hfill \\ \hfill 2\,{\mathrm{cos}}^{2}\left(\theta\right) -\mathrm{cos}\left(\theta\right) -1& =& 0\hfill \\ \hfill \left(2\,\mathrm{cos}\left(\theta\right) +1\right)\left(\mathrm{cos}\left(\theta\right) -1\right)& =& 0\hfill \\ \hfill 2\,\mathrm{cos}\left(\theta\right) +1& =& 0\hfill \\ \hfill \mathrm{cos}\left(\theta\right) & =& -\frac{1}{2}\hfill \\ & & \\ \hfill \mathrm{cos}\left(\theta\right) -1& =& 0\hfill \\ \hfill \mathrm{cos}\left(\theta\right) & =& 1\hfill \end{array}[/latex]
So, if [latex]\mathrm{cos}\left(\theta\right) =-\frac{1}{2},[/latex] then [latex]\theta =\frac{2\pi }{3}±2\pi k\,[/latex] and [latex]\theta =\frac{4\pi }{3}±2\pi k;[/latex] if [latex]\mathrm{cos}\left(\theta\right) =1,[/latex] then [latex]\theta =0±2\pi k.[/latex]
EXAMPLE 14
Solving an Equation Using an Identity
Solve the equation exactly using an identity: [latex]3\,\mathrm{cos}\left(\theta\right) +3=2\;{\mathrm{sin}}^{2}\left(\theta\right),\; \; \; 0\le \theta \lt2\pi.[/latex]
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Solution
Our solutions are [latex]\theta =\frac{2\pi }{3},\,\frac{4\pi }{3},\,\pi.[/latex]
Section 3.2 Exercises
[Answers to odd problem numbers are provided at the end of the problem set. Just scroll down!]
Verbal
1. Will there always be solutions to trigonometric function equations? If not, describe an equation that would not have a solution. Explain why or why not.
2. When solving a trigonometric equation involving more than one trig function, do we always want to try to rewrite the equation so it is expressed in terms of one trigonometric function? Why or why not?
3. When solving linear trig equations in terms of only sine or cosine, how do we know whether there will be solutions?
Algebraic
For the following exercises, find all solutions exactly on the interval[latex]\,0\le \theta \lt 2\pi .[/latex]
4. [latex]2\,\mathrm{sin}\left(\theta\right) =-\sqrt{2}[/latex]
5. [latex]2\,\mathrm{sin}\left(\theta\right) =\sqrt{3}[/latex]
6. [latex]2\,\mathrm{cos}\left(\theta\right) =1[/latex]
7. [latex]2\,\mathrm{cos}\left(\theta\right) =-\sqrt{2}[/latex]
8. [latex]\mathrm{tan}\left(\theta\right) =-1[/latex]
9. [latex]\mathrm{tan}\left(x\right)=1[/latex]
10. [latex]\mathrm{cot}\left(x\right)+1=0[/latex]
11. [latex]4\,{\mathrm{sin}}^{2}\left(x\right)-2=0[/latex]
12. [latex]{\mathrm{csc}}^{2}\left(x\right)-4=0[/latex]
For the following exercises, solve exactly on[latex]\,\left[0,2\pi \right).[/latex]
13. [latex]2\,\mathrm{cos}\left(\theta\right) =\sqrt{2}[/latex]
14. [latex]2\,\mathrm{cos}\left(\theta\right) =-1[/latex]
15. [latex]2\,\mathrm{sin}\left(\theta\right) =-1[/latex]
16. [latex]2\,\mathrm{sin}\left(\theta\right) =-\sqrt{3}[/latex]
17. [latex]2\,\mathrm{sin}\left(3\theta \right)=1[/latex]
18. [latex]2\,\mathrm{sin}\left(2\theta \right)=\sqrt{3}[/latex]
19. [latex]2\,\mathrm{cos}\left(3\theta \right)=-\sqrt{2}[/latex]
20. [latex]\mathrm{cos}\left(2\theta \right)=-\frac{\sqrt{3}}{2}[/latex]
21. [latex]2\,\mathrm{sin}\left(\pi \theta \right)=1[/latex]
22. [latex]2\,\mathrm{cos}\left(\frac{\pi }{5}\theta \right)=\sqrt{3}[/latex]
For the following exercises, find all exact solutions on[latex]\,\left[0,2\pi \right).[/latex]
23. [latex]\mathrm{sec}\left(x\right)\mathrm{sin}\left(x\right)-2\,\mathrm{sin}\left(x\right)=0[/latex]
24. [latex]\mathrm{tan}\left(x\right)-2\,\mathrm{sin}\left(x\right)\mathrm{tan}\left(x\right)=0[/latex]
25. [latex]2\,{\mathrm{cos}}^{2}\left(t\right)+\mathrm{cos}\left(t\right)=1[/latex]
26. [latex]2\,{\mathrm{tan}}^{2}\left(t\right)=3\,\mathrm{sec}\left(t\right)[/latex]
27. [latex]2\,\mathrm{sin}\left(x\right)\mathrm{cos}\left(x\right)-\mathrm{sin}\left(x\right)+2\,\mathrm{cos}\left(x\right)-1=0[/latex]
28. [latex]{\mathrm{cos}}^{2}\left(\theta\right) =\frac{1}{2}[/latex]
29. [latex]{\mathrm{sec}}^{2}\left(x\right)=1[/latex]
30. [latex]{\mathrm{tan}}^{2}\left(x\right)=-1+2\,\mathrm{tan}\left(-x\right)[/latex]
31. [latex]8\,{\mathrm{sin}}^{2}\left(x\right)+6\,\mathrm{sin}\left(x\right)+1=0[/latex]
32. [latex]{\mathrm{tan}}^{5}\left(x\right)=\mathrm{tan}\left(x\right)[/latex]
For the following exercises, solve with the methods shown in this section exactly on the interval[latex]\,\left[0,2\pi \right).[/latex]
33. [latex]\mathrm{sin}\left(3x\right)\mathrm{cos}\left(6x\right)-\mathrm{cos}\left(3x\right)\mathrm{sin}\left(6x\right)=-0.9[/latex]
34. [latex]\mathrm{sin}\left(6x\right)\mathrm{cos}\left(11x\right)-\mathrm{cos}\left(6x\right)\mathrm{sin}\left(11x\right)=-0.1[/latex]
35. [latex]\mathrm{cos}\left(2x\right)\mathrm{cos}\left(x\right)+\mathrm{sin}\left(2x\right)\mathrm{sin}\left(x\right)=1[/latex]
36. [latex]6\,\mathrm{sin}\left(2t\right)+9\,\mathrm{sin}\left(t\right)=0[/latex]
37. [latex]9\,\mathrm{cos}\left(2\theta \right)=9\,{\mathrm{cos}}^{2}\left(\theta\right) -4[/latex]
38. [latex]\mathrm{sin}\left(2t\right)=\mathrm{cos}\left(t\right)[/latex]
39. [latex]\mathrm{cos}\left(2t\right)=\mathrm{sin}\left(t\right)[/latex]
40. [latex]\mathrm{cos}\left(6x\right)-\mathrm{cos}\left(3x\right)=0[/latex]
For the following exercises, solve exactly on the interval[latex]\,\left[0,2\pi \right)[/latex]. Use the quadratic formula if the equations do not factor.
41. [latex]{\mathrm{tan}}^{2}\left(x\right)-\sqrt{3}\,\mathrm{tan}\left(x\right)=0[/latex]
42. [latex]{\mathrm{sin}}^{2}\left(x\right)+\mathrm{sin}\left(x\right)-2=0[/latex]
43. [latex]{\mathrm{sin}}^{2}\left(x\right)-2\,\mathrm{sin}\left(x\right)-4=0[/latex]
44. [latex]5\,{\mathrm{cos}}^{2}\left(x\right)+3\,\mathrm{cos}\left(x\right)-1=0[/latex]
45. [latex]3\,{\mathrm{cos}}^{2}\left(x\right)-2\,\mathrm{cos}\left(x\right)-2=0[/latex]
46. [latex]5\,{\mathrm{sin}}^{2}\left(x\right)+2\,\mathrm{sin}\left(x\right)-1=0[/latex]
47. [latex]{\mathrm{tan}}^{2}\left(x\right)+5\mathrm{tan}\left(x\right)-1=0[/latex]
48. [latex]{\mathrm{cot}}^{2}\left(x\right)=-\mathrm{cot}\left(x\right)[/latex]
49. [latex]-{\mathrm{tan}}^{2}\left(x\right)-\mathrm{tan}\left(x\right)-2=0[/latex]
For the following exercises, find exact solutions on the interval[latex]\,\left[0,2\pi \right)[/latex]. Look for opportunities to use trigonometric identities.
50. [latex]{\mathrm{sin}}^{2}\left(x\right)-{\mathrm{cos}}^{2}\left(x\right)-\mathrm{sin}\left(x\right)=0[/latex]
51. [latex]{\mathrm{sin}}^{2}\left(x\right)+{\mathrm{cos}}^{2}\left(x\right)=0[/latex]
52. [latex]\mathrm{sin}\left(2x\right)-\mathrm{sin}\left(x\right)=0[/latex]
53. [latex]\mathrm{cos}\left(2x\right)-\mathrm{cos}\left(x\right)=0[/latex]
54. [latex]\frac{2\,\mathrm{tan}\left(x\right)}{2-{\mathrm{sec}}^{2}\left(x\right)}-{\mathrm{sin}}^{2}\left(x\right)={\mathrm{cos}}^{2}\left(x\right)[/latex]
55. [latex]1-\mathrm{cos}\left(2x\right)=1+\mathrm{cos}\left(2x\right)[/latex]
56. [latex]{\mathrm{sec}}^{2}\left(x\right)=7[/latex]
57. [latex]10\,\mathrm{sin}\left(x\right)\,\mathrm{cos}\left(x\right)=6\,\mathrm{cos}\left(x\right)[/latex]
58. [latex]-3\,\mathrm{sin}\left(t\right)=15\,\mathrm{cos}\left(t\right)\,\mathrm{sin}\left(t\right)[/latex]
59. [latex]4\,{\mathrm{cos}}^{2}\left(x\right)-4=15\,\mathrm{cos}\left(x\right)[/latex]
60. [latex]8\,{\mathrm{sin}}^{2}\left(x\right)+6\,\mathrm{sin}\left(x\right)+1=0[/latex]
61. [latex]8\,{\mathrm{cos}}^{2}\left(\theta\right) =3-2\,\mathrm{cos}\left(\theta\right)[/latex]
62. [latex]6\,{\mathrm{cos}}^{2}\left(x\right)+7\,\mathrm{sin}\left(x\right)-8=0[/latex]
63. [latex]12\,{\mathrm{sin}}^{2}\left(t\right)+\mathrm{cos}\left(t\right)-6=0[/latex]
64. [latex]\mathrm{tan}\left(x\right)=3\,\mathrm{sin}\left(x\right)[/latex]
65. [latex]{\mathrm{cos}}^{3}\left(t\right)=\mathrm{cos}\left(t\right)[/latex]
Answers to Section 3.2 Odd Problems
1. There will not always be solutions to trigonometric function equations. For a basic example,[latex]\,\mathrm{cos}\left(x\right)=-5.[/latex]
3. If the sine or cosine function has a coefficient of one, isolate the term on one side of the equals sign. If the number it is set equal to has an absolute value less than or equal to one, the equation has solutions, otherwise it does not. If the sine or cosine does not have a coefficient equal to one, still isolate the term but then divide both sides of the equation by the leading coefficient. Then, if the number it is set equal to has an absolute value greater than one, the equation has no solution.
5. [latex]\frac{\pi }{3},\frac{2\pi }{3}[/latex]
7. [latex]\frac{3\pi }{4},\frac{5\pi }{4}[/latex]
9. [latex]\frac{\pi }{4},\frac{5\pi }{4}[/latex]
11. [latex]\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}[/latex]
13. [latex]\frac{\pi }{4},\frac{7\pi }{4}[/latex]
15. [latex]\frac{7\pi }{6},\frac{11\pi }{6}[/latex]
17. [latex]\frac{\pi }{18},\frac{5\pi }{18},\frac{13\pi }{18},\frac{17\pi }{18},\frac{25\pi }{18},\frac{29\pi }{18}[/latex]
19. [latex]\frac{3\pi }{12},\frac{5\pi }{12},\frac{11\pi }{12},\frac{13\pi }{12},\frac{19\pi }{12},\frac{21\pi }{12}[/latex]
21. [latex]\frac{1}{6},\frac{5}{6},\frac{13}{6},\frac{17}{6},\frac{25}{6},\frac{29}{6},\frac{37}{6}[/latex]
23. [latex]0,\frac{\pi }{3},\pi ,\frac{5\pi }{3}[/latex]
25. [latex]\frac{\pi }{3},\pi ,\frac{5\pi }{3}[/latex]
27. [latex]\frac{\pi }{3},\frac{3\pi }{2},\frac{5\pi }{3}[/latex]
29. [latex]0,\pi[/latex]
31. [latex]\pi -{\mathrm{sin}}^{-1}\left(-\frac{1}{4}\right),\frac{7\pi }{6},\frac{11\pi }{6},2\pi +{\mathrm{sin}}^{-1}\left(-\frac{1}{4}\right)[/latex]
33. [latex]\frac{1}{3}\left({\mathrm{sin}}^{-1}\left(\frac{9}{10}\right)\right),\frac{\pi }{3}-\frac{1}{3}\left({\mathrm{sin}}^{-1}\left(\frac{9}{10}\right)\right),\frac{2\pi }{3}+\frac{1}{3}\left({\mathrm{sin}}^{-1}\left(\frac{9}{10}\right)\right),\pi -\frac{1}{3}\left({\mathrm{sin}}^{-1}\left(\frac{9}{10}\right)\right),\frac{4\pi }{3}+\frac{1}{3}\left({\mathrm{sin}}^{-1}\left(\frac{9}{10}\right)\right),\frac{5\pi }{3}-\frac{1}{3}\left({\mathrm{sin}}^{-1}\left(\frac{9}{10}\right)\right)[/latex]
35. [latex]0[/latex]
37. [latex]\frac{\pi }{6},\frac{5\pi }{6},\frac{7\pi }{6},\frac{11\pi }{6}[/latex]
39. [latex]\frac{3\pi }{2},\frac{\pi }{6},\frac{5\pi }{6}[/latex]
41. [latex]0,\frac{\pi }{3},\pi ,\frac{4\pi }{3}[/latex]
43. There are no solutions.
45. [latex]{\mathrm{cos}}^{-1}\left(\frac{1}{3}\left(1-\sqrt{7}\right)\right),2\pi -{\mathrm{cos}}^{-1}\left(\frac{1}{3}\left(1-\sqrt{7}\right)\right)[/latex]
47. [latex]{\mathrm{tan}}^{-1}\left(\frac{1}{2}\left(\sqrt{29}-5\right)\right),\pi +{\mathrm{tan}}^{-1}\left(\frac{1}{2}\left(-\sqrt{29}-5\right)\right),\pi +{\mathrm{tan}}^{-1}\left(\frac{1}{2}\left(\sqrt{29}-5\right)\right),2\pi +{\mathrm{tan}}^{-1}\left(\frac{1}{2}\left(-\sqrt{29}-5\right)\right)[/latex]
49. There are no solutions.
51. There are no solutions.
53. [latex]0,\frac{2\pi }{3},\frac{4\pi }{3}[/latex]
55. [latex]\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}[/latex]
57. [latex]{\mathrm{sin}}^{-1}\left(\frac{3}{5}\right),\frac{\pi }{2},\pi -{\mathrm{sin}}^{-1}\left(\frac{3}{5}\right),\frac{3\pi }{2}[/latex]
59. [latex]{\mathrm{cos}}^{-1}\left(-\frac{1}{4}\right),2\pi -{\mathrm{cos}}^{-1}\left(-\frac{1}{4}\right)[/latex]
61. [latex]\frac{\pi }{3},{\mathrm{cos}}^{-1}\left(-\frac{3}{4}\right),2\pi -{\mathrm{cos}}^{-1}\left(-\frac{3}{4}\right),\frac{5\pi }{3}[/latex]
63. [latex]{\mathrm{cos}}^{-1}\left(\frac{3}{4}\right),{\mathrm{cos}}^{-1}\left(-\frac{2}{3}\right),2\pi -{\mathrm{cos}}^{-1}\left(-\frac{2}{3}\right),2\pi -{\mathrm{cos}}^{-1}\left(\frac{3}{4}\right)[/latex]
65. [latex]0,\frac{\pi }{2},\pi ,\frac{3\pi }{2}[/latex]
