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Precalculus : Solving Trigonometric Equations and Inequalities

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Example Question #1 : Solving Trigonometric Equations And Inequalities

Use trigonometric identities to solve the following equation for $\theta$ :

$tan^2\theta + sec^2\theta = 1$

Possible Answers:

$\theta= n\pi$

$\theta = \pi/3+ n$

$\theta = \pi/2+ n\pi$

$\theta = \pi/4+ n\pi$

Correct answer:

$\theta= n\pi$

Explanation:

Use the trigonometric identities to switch sec into terms of tan:

$tan^2\theta + sec^2\theta = tan^2\theta + (tan^2 \theta + 1) = 2tan^2\theta + 1 = 1$

hence,

$2tan^2\theta = 0$

So we have $\tan \theta = 0$ , making $\theta= n \pi$

Therefore the solution is $\theta = n \pi$ for n being any integer.

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Example Question #2 : Solve Trigonometric Equations And Inequalities

Which of the following is not a solution to $\small 2sin^2(x)-3sin(x)=-1$ for $\small 0\leq x<2\pi$

Possible Answers:

$\small \frac{\pi}{6}$

$\small \frac{\pi}{2}$

$\small \frac{11\pi}{6}$

$\small \frac{5\pi}{6}$

Correct answer:

$\small \frac{11\pi}{6}$

Explanation:

We begin by setting the right side of the equation equal to 0.

$\small 2sin^2(x)-3sin(x)+1=0$

The equation might be easier to factor using the following substitution.

$\small w=sin(x)$

This gives the following

$\small 2w^2-3w+1=0$

This can be factored as follows

$\small \small (2w-1)(w-1)=0$

Therefore

$\small w=\frac{1}{2}\;or\;w=1$

Replacing our substitution therefore gives

$\small sin(x)=\frac{1}{2}\;or\;sin(x)=1$

Within our designated domain, we get three answers between our two equations.

$\small sin(x)=\frac{1}{2}$ when $\small x=\frac{\pi}{6}\;or\; \frac{5\pi}{6}$

$\small sin(x)=1$ when $\small x=\frac{\pi}{2}$

Therefore, the only choice that isn't correct is $\small \frac{11\pi}{6}$

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Example Question #3 : Solve Trigonometric Equations And Inequalities

Find one possible value of .

tan(5x)+17=-55

Possible Answers:

$-446.02^{\circ}$

$17.84^{\circ}$

$446.02^{\circ}$

$-17.84^{\circ}$

$0^{\circ}$

Correct answer:

$-17.84^{\circ}$

Explanation:

Begin by isolating the tangent side of the equation:

tan(5x)+17=-55

tan(5x)=-72

Next, take the inverse tangent of both sides:

$5x=tan^{-1} (-72)$

Divide by five to get the final answer:

$x=\frac{tan^{-1} (-72)}{5}=-17.84^{\circ}$

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Example Question #4 : Solve Trigonometric Equations And Inequalities

Use trigonometric identities to solve for the angle value.

$\tan^2\theta\sin^2\theta+\tan^2\theta\cos^2\theta+1=13$

Possible Answers:

$\theta\approx73.898$

$\theta\approx85.601$

$\theta\approx53.565$

$\theta\approx74.499$

Correct answer:

$\theta\approx73.898$

Explanation:

There are two ways to solve this problem. The first involves two trigonometric identities:

$\tan^2\theta\sin^2\theta+\tan^2\theta\cos^2\theta+1=13\\ \implies\tan^2\theta(\sin^2\theta+\cos^2\theta)+1=13\\ \implies\tan^2\theta+1=13\implies\sec^2\theta=13\\ \implies\sec\theta=\sqrt13\implies\cos\theta=\frac{\sqrt13}{13}\\ \implies \theta=\cos^{-1}\frac{\sqrt13}{13}}\approx73.898}$

The second method allows us to only use the first trigonometric identity:

$\tan^2\theta\sin^2\theta+\tan^2\theta\cos^2\theta+1=13\\ \implies\tan^2\theta(\sin^2\theta+\cos^2\theta)+1=13\\ \implies\tan^2\theta+1=13\implies\tan^2\theta=12\\ \implies\tan\theta=2\sqrt3\implies\theta=\tan^{-1}2\sqrt3\approx73.898}$

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Example Question #5 : Solve Trigonometric Equations And Inequalities

Use trigonometric identities to solve the equation for the angle value.

$4\cos^2\theta+\cos{2\theta}+\sin^2\theta=3$

Possible Answers:

$\theta=\frac{\pi}{6}$

$\theta\approx39.232$

$\theta\approx78.463$

$\theta=\frac{\pi}{2}$

Correct answer:

$\theta\approx39.232$

Explanation:

The simplest way to solve this problem is using the double angle identity for cosine. $\cos2\theta=\cos^2\theta-\sin^2\theta$

Substituting this value into the original equation gives us:

$4\cos^2\theta+\cos^2\theta-\sin^2\theta+\sin^2\theta=3\\ \implies5\cos^2\theta=3\implies\cos^2\theta=\frac{3}{5}\\ \cos\theta=\frac{\sqrt15}{5}\implies\theta=\cos^{-1}\frac{\sqrt15}{5}}\approx39.232$

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Example Question #3 : Solving Trigonometric Equations And Inequalities

According to the trigonometric identities, 1+cot^2(x)=?

Possible Answers:

$\frac{sin(x)}{cos(x)}$

tan(x)

$\frac{1}{tan(x)}$

csc^2(x)

Correct answer:

csc^2(x)

Explanation:

The trigonometric identity 1+cot^2(x)=csc^2(x) , is an important identity to memorize.

Some other identities that are important to know are:

$\frac{sin(x)}{cos(x)}=tan(x)$

$cot(x)=\frac{1}{tan(x)}$

sin^2(x)+cos^2(x)=1

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Example Question #1 : Solve Trigonometric Equations And Inequalities In Quadratic Form

If $\theta$ exists in the domain from $\left[0,\frac{\pi}{2}\right]$ , solve the following: $0=1-\cos^2\theta$

Possible Answers:

$0,\frac{\pi}{2}$

$\textup{No solution.}$

$\pi$

$0,\pi$

Correct answer:

Explanation:

Factorize $0=1-cos^2\theta$ .

$1-cos^2\theta= (1+cos(\theta))(1-cos(\theta))$

Set both terms equal to zero and solve.

$1+cos(\theta)=0$

$cos(\theta)=-1$

$\theta=\pi$

This value is not within the $[0,\frac{\pi}{2}]$ domain.

$1-cos(\theta)=0$

$cos(\theta)=1$

$\theta=0$

This is the only correct value in the $[0,\frac{\pi}{2}]$ domain.

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Example Question #2 : Solve Trigonometric Equations And Inequalities In Quadratic Form

Solve for $\theta$ in the equation $sin^2(\theta)-2sin(\theta)+1=0$ on the interval $0\leq\theta\leq2\pi$ .

Possible Answers:

$\theta=\frac{\pi}{2}$

$\theta=0$

$\theta=sin^{-1}(1)$

$\theta=\frac{3\pi}{2}$

$\theta=\frac{3\pi}{4}$

Correct answer:

$\theta=\frac{\pi}{2}$

Explanation:

If you substitute $x=sin(\theta)$ you obtain a recognizable quadratic equation which can be solved for ,

(x-1)^2=0 .

Then we can plug back into our equation and use the unit circle to find that

$\theta=\frac{\pi}{2}$ .

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Example Question #3 : Solve Trigonometric Equations And Inequalities In Quadratic Form

Given that theta exists from $[0,2\pi]$ , solve: $2cos^2\theta=cos(\theta)$

Possible Answers:

$\frac{\pi}{3},\frac{\pi}{2},\frac{3\pi}{2}, \frac{5\pi}{3}$

$\textup{None of the given answers.}$

$\frac{\pi}{2}$

$\frac{\pi}{2},\frac{3\pi}{2}$

$\frac{\pi}{3}, \frac{5\pi}{3}$

Correct answer:

$\frac{\pi}{3},\frac{\pi}{2},\frac{3\pi}{2}, \frac{5\pi}{3}$

Explanation:

In order to solve $2cos^2\theta=cos(\theta)$ appropriately, do not divide $cos(\theta)$ on both sides. The effect will eliminate one of the roots of this trig function.

Substract $cos(\theta)$ from both sides.

$2cos^2\theta-cos(\theta)=0$

Factor the left side of the equation.

$cos(\theta)[2cos(\theta)-1]=0$

Set each term equal to zero, and solve for theta with the restriction $[0,2\pi]$ .

$cos(\theta)=0$

$\theta = \frac{\pi}{2},\frac{3\pi}{2}$

$2cos(\theta)-1=0$

$2cos(\theta)=1$

$cos(\theta)=\frac{1}{2}$

$\theta= \frac{\pi}{3}, \frac{5\pi}{3}$

The correct answer is:

$\theta=\frac{\pi}{2},\frac{3\pi}{2},\frac{\pi}{3}, \frac{5\pi}{3}$

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Example Question #1 : Solve Trigonometric Equations And Inequalities In Quadratic Form

Solve $\sin^2(x) + 4 = 0$ for $0^{\circ}\leq x\leq 360^{\circ}$

Possible Answers:

$x = 90^{\circ}$

x = 0

There is no solution.

x = -2, x = 2

Correct answer:

There is no solution.

Explanation:

By subtracting from both sides of the original equation, we get $\sin^2(x) = -4$ . We know that the square of a trigonometric identity cannot be negative, regardless of the input, so there can be no solution.

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Precalculus : Solving Trigonometric Equations and Inequalities

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1 : Solving Trigonometric Equations And Inequalities

Example Question #2 : Solve Trigonometric Equations And Inequalities

Example Question #3 : Solve Trigonometric Equations And Inequalities

Example Question #4 : Solve Trigonometric Equations And Inequalities

Example Question #5 : Solve Trigonometric Equations And Inequalities

Example Question #3 : Solving Trigonometric Equations And Inequalities

Example Question #1 : Solve Trigonometric Equations And Inequalities In Quadratic Form

Example Question #2 : Solve Trigonometric Equations And Inequalities In Quadratic Form

Example Question #3 : Solve Trigonometric Equations And Inequalities In Quadratic Form

Example Question #1 : Solve Trigonometric Equations And Inequalities In Quadratic Form

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