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Trigonometry : Solving Trigonometric Equations

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Example Questions

Example Question #185 : Trigonometry

For this question, we will denote by max the maximum value of the function and min the minimum value of the function.

What is the maximum and minimum values of

cos^4x+sin^3x, where is a real number.

Possible Answers:

max=0, min=-1

max=1, min=0

max=2, min=-2

max=2, min=-1

max=1, min=-1

Correct answer:

max=2, min=-1

Explanation:

To find the maximum and the minimum , we can view the above function as

a system where $-1\le cosx\le 1$ and $-1\le sinx\le 1$ . Using these two conditions we find the maximum and the minimum.

$0\le cos^2x\le 1$ means also that $0\le cos^4x\le 1$ ( $\star$ ) We also have:

$-1\le sinx\le 1$ implies that :

$-1\le sin^{3}x\le 1$ ( $\star \star$ ) Therefore we have by adding ( $\star$ ) and( $\star \star$ )

$-1\le cos^4x+sin^3x\le 2$

This means that max=2 and min=-1

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Example Question #1 : Solving Trigonometric Equations

Find the values of that satisfy the following system:

$\left\{\begin{matrix} 2cosx\le \sqrt{2}\\2sinx\le \sqrt{2} \\ cosx< sinx\end{matrix}\right.$

where is assumed to be $[0,\frac{\pi}{2}]$

Possible Answers:

This system does not have a solution.

$[\frac{\pi}{4},\frac{\pi}{2}]$

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

$[-\frac{\pi}{4},\frac{\pi}{2}]$

$[0,\frac{\pi}{4}]$

Correct answer:

This system does not have a solution.

Explanation:

We can write the system in the equivalent form:

$cosx\le \frac{\sqrt{2}}{2}$

$sinx\le \frac{\sqrt{2}}{2}$

cosx< sinx

The solution to the first equation is $x\in \[0,\frac{\pi}{4}]$

cosx< sinx means that $x\in (\frac{\pi}{4},\frac{\pi}{2}]$

This means that there is no x that satisfies the system.

Therefore there is no x that solves the 3 inequalities simultaneously.

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Example Question #1 : Systems Of Trigonometric Equations

Which of the following systems of trigonometric equations have a solution with an -coordinate of $\frac{\pi}{4}$ ?

Possible Answers:

y=sin(x)

y=cos(7x)

y=sin(x)

y=cos(3x)

More than one of these answers has a solutions at $x=\frac{\pi}{4}$ .

y=sin(x)

y=cos(5x)

Correct answer:

y=sin(x)

y=cos(7x)

Explanation:

The solution to the correct answer would be $\left(\frac{\pi}{4}, \frac{1}{\sqrt2}\right)$ .

For all of the other answers, plugging in $\frac{\pi}{4}$ for the second equation gives a y value of $-\frac{1}{\sqrt2}$ .

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Example Question #1 : Solving Trigonometric Equations

Solve the system for $\theta$ :

$\begin{matrix} y = 6 \sin \theta - 1 \\ y = \sin ^2 \theta + 4 \end{matrix}$

Possible Answers:

no solution

$\pi$

$2 \pi$

$\frac{ \pi }{2 }$

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

Correct answer:

$\frac{ \pi }{2 }$

Explanation:

First, set both equations equal to each other:

$6 \sin \theta - 1 = \sin ^2 \theta + 4$ subtract $6 \sin \theta$ from both sides

$-1 = \sin ^2 \theta - 6 \sin \theta + 4$ add 1 to both sides

$0 = \sin ^2 \theta - 6 \sin \theta + 5$

Now we can solve this as a quadratic equation, where "x" is $\sin \theta$ . Using the quadratic formula:

$\sin \theta = \frac{ 6 \pm \sqrt{ 36 - 4(1)(5)}}{2} = \frac{6 \pm \sqrt{16 }}{2} = \frac{ 6 \pm 4 }{2}$

This gives us 2 potential solutions for $\sin \theta$ :

$\sin \theta = \frac{10}{2} = 5$ the sine of an angle cannot be greater than 1

$\sin \theta = \frac{2}{2 } = 1$

$\theta = \sin ^ {-1} (1) = \frac{ \pi }{2}$

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Example Question #1 : Systems Of Trigonometric Equations

Solve this system for $\theta$ :

$\begin{matrix} y = 5 \sin (\frac{ \theta }{2}) + 3 \\ y = 3 \sin (\frac{\theta }{2} ) + 2 \end{matrix}$

Possible Answers:

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

$\frac{ 11 \pi }{6} , \frac{ 7 \pi }{6}$

$\frac{ 11 \pi }{3} , \frac{ 7 \pi }{3}$

$\frac{ 11 \pi }{12} , \frac{ 7 \pi }{12}$

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

Correct answer:

$\frac{ 11 \pi }{3} , \frac{ 7 \pi }{3}$

Explanation:

First, set the two equations equal to each other

$5 \sin (\frac{ \theta }{2}) + 3 = 3 \sin (\frac{ \theta }{2} ) + 2$ subtract the sine term from the right

$2 \sin (\frac{ \theta } { 2 }) + 3 = 2$ subtract 3 from both sides

$2 \sin (\frac{ \theta }{2} ) = -1$ divide by 2

$\sin (\frac{ \theta }{2} ) = -\frac{1}{2}$

$\frac{ \theta } {2} = \sin ^{-1} (- \frac{1}{2}) = \frac{ 11 \pi }{6} , \frac{ 7 \pi }{6}$ multiply by 2

$\theta = \frac{ 11 \pi }{3} , \frac{ 7 \pi }{3}$

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Example Question #2 : Systems Of Trigonometric Equations

Solve this system for $\theta$ :

$y = \sin ^ 2 \theta + \cos \theta$

$y = \cos \theta + \frac{1}{2}$

Possible Answers:

no solution

$\frac{ \pi }{4} , \frac{ 3 \pi }{4}$

$\frac{ \pi }{4} , \frac{ 3 \pi }{4} ,\frac{ 5 \pi }{4} , \frac{ 7 \pi }{4 }$

$\frac{ \pi }{4} , \frac{ 7 \pi }{4}$

$\frac{ 3 \pi }{4} , \frac{ 5 \pi }{4}$

Correct answer:

$\frac{ \pi }{4} , \frac{ 3 \pi }{4} ,\frac{ 5 \pi }{4} , \frac{ 7 \pi }{4 }$

Explanation:

Set the two equations equal to each other

$\sin ^2 \theta + \cos \theta = \cos \theta + \frac{ 1}{2}$ subtract cos from both sides

$\sin ^ 2 \theta = \frac{ 1}{2 }$ take the square root of both sides

$\sin \theta = \pm \frac{ 1}{ \sqrt2}$

$\theta = \sin ^{-1} ( \pm \frac{ 1} {\sqrt2}) = \frac{ \pi }{4} , \frac{ 3 \pi }{4} , \frac{ 5 \pi }{4} , \frac{ 7 \pi }{4}$

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Example Question #31 : Trigonometric Equations

Solve this system for $\theta$ :

$\\ y = \cos ^2 \theta + \frac{1}{2} \\ y = \sin ^2 \theta$

Possible Answers:

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

$\frac{4 \pi }{3} , \frac{ 2 \pi }{3}$

$\frac{ 7 \pi }{6} , \frac{ 11 \pi }{6}$

$\frac{ 2 \pi }{3 } , \frac{4 \pi }{3}$

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

Correct answer:

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

Explanation:

Set both equations equal to each other:

$\cos ^2 \theta + \frac{1}{2} = \sin ^2 \theta$ subtract $\frac{1}{2}$ from both sides

$\cos ^2 \theta = \sin ^2 \theta - \frac{1}{2}$ subtract $\sin ^2 \theta$ from both sides

$\cos ^2 \theta - \sin ^2 \theta = - \frac{1}{2}$

We can re-write the left side using a trigonometric identity

$\cos ( 2 \theta ) = -\frac{1}{2 }$ take the inverse cosine

$2 \theta = \cos ^ {-1}\left(-\frac{1}{2} \right)$

$2 \theta = \frac{ 2 \pi }{3} , \frac{4 \pi }{3 }$ divide by 2

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

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Example Question #1 : Solving Trigonometric Equations

Solve this system for $\theta$ :

$\\y = 4 \cos ^ 2 \theta + \cos \theta - 5 \\ y = \cos \theta - 2$

Possible Answers:

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

$\frac{ \pi }{3} , \frac{ 11 \pi }{3}$

$\frac{ \pi }{6} , \frac {5 \pi }{6} , \frac{ 7 \pi }{6} , \frac{ 11 \pi }{6}$

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

no solution

Correct answer:

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

Explanation:

Set the two equations equal to each other:

$4 \cos ^2 \theta + \cos \theta - 5 = \cos \theta - 2$ subtract $\cos \theta$ from both sides

$4 \cos ^ 2 \theta - 5 = -2$ add 5 to both sides

$4 \cos ^ 2 \theta = 3$ divide both sides by 4

$\cos ^ 2 \theta = \frac{ 3}{4}$ take the square root of both sides

$\cos \theta = \sqrt{\frac{ 3}{4} } = \frac{ \sqrt{3} }{\sqrt{4}} = \pm \frac{ \sqrt{3}}{2}$

$\theta = \cos ^ {-1} \left( \pm \frac{ \sqrt 3}{2}\right)$

$\theta = \frac{ \pi }{6} , \frac{ 11 \pi }{6}$

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Example Question #2 : Solving Trigonometric Equations

Solve the following system:

Screen shot 2020 05 21 at 10.30.36 am

Possible Answers:

$-\pi$

The system does not have a solution.

$\pi$

$\frac{-\pi}{3}$

$\frac{\pi}{3}$

Correct answer:

The system does not have a solution.

Explanation:

A number x is a solution if it satisfies both equations.

We note first we can write the first equation in the form :

$cosx\ge \frac{4}{3}$

We know that $cosx\le 1$ for all reals. This means that there is no x that

satisifies the first inequality. This shows that the system cannot satisfy both equations since it does not satisfy one of them. This shows that our system does not have a solution.

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Example Question #4 : Systems Of Trigonometric Equations

Solve this system for $\theta$ :

$\begin{matrix} y = \sin ^2 \theta + 1 \\ y = \cos \theta + 2 \end{matrix}$

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

First, set both equations equal to each other:

$\sin ^2 \theta + 1 = \cos \theta + 2$ subtract $\cos \theta$ from both sides

$\sin ^2 \theta - \cos \theta + 1 = 2$

Using a trigonometric identity, we can re-write $\sin^2 \theta$ as $1 - \cos ^2 \theta$ :

$1 - \cos^2 \theta - \cos \theta + 1 =2$ combine like terms

$- \cos ^2 \theta - \cos \theta + 2 = 2$ subtract 2 from both sides

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

We can solve for $\cos \theta$ using the quadratic formula:

$\cos \theta = \frac{ 1 \pm \sqrt{ 1 - 4(-1)(0)}}{2(-1)} = \frac{ 1 \pm 1}{ -2 }$

This gives us 2 possible values for cosine

$\cos \theta = \frac{ 1 + 1 }{ -2 } = \frac{ 2 }{ - 2 } = -1$

$\theta = \cos ^ {-1} ( -1 ) = \pi$

$\cos \theta = \frac{ 1 - 1 }{-2 } = \frac{ 0 }{-2} = 0$

$\theta = \cos ^ {-1} (0) = \frac{ \pi }{2} , \frac{ 3 \pi }{2}$

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Trigonometry : Solving Trigonometric Equations

Study concepts, example questions & explanations for Trigonometry

All Trigonometry Resources

Example Questions

Example Question #185 : Trigonometry

Example Question #1 : Solving Trigonometric Equations

Example Question #1 : Systems Of Trigonometric Equations

Example Question #1 : Solving Trigonometric Equations

Example Question #1 : Systems Of Trigonometric Equations

Example Question #2 : Systems Of Trigonometric Equations

Example Question #31 : Trigonometric Equations

Example Question #1 : Solving Trigonometric Equations

Example Question #2 : Solving Trigonometric Equations

Example Question #4 : Systems Of Trigonometric Equations

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