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Precalculus : Solve 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:

$\theta= n\pi$

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

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

$\theta = \pi/3+ n$

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 #1 : Solving 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:

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

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

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

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

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 : Solving 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 : Solving 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:

$\theta\approx74.499$

$\theta\approx85.601$

$\theta\approx53.565$

$\theta\approx73.898$

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 : Solving 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:

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

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

$\theta\approx78.463$

$\theta\approx39.232$

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 #6 : Solving Trigonometric Equations And Inequalities

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

Possible Answers:

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

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

csc^2(x)

tan(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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Precalculus : Solve 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 #1 : Solving Trigonometric Equations And Inequalities

Example Question #3 : Solving Trigonometric Equations And Inequalities

Example Question #4 : Solving Trigonometric Equations And Inequalities

Example Question #5 : Solving Trigonometric Equations And Inequalities

Example Question #6 : Solving Trigonometric Equations And Inequalities

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