Systems of Trigonometric Equations - Trigonometry

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Question

Solve this system for $\theta$ :

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

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

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Answer

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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Question

Solve the following system:

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Answer

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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Question

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

where is a real number.

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Answer

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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Question

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

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Answer

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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Question

Solve the system for $\theta$ :

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

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Answer

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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Question

Solve this system for $\theta$ :

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

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Answer

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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Question

Solve this system for $\theta$ :

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

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Answer

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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Question

Solve this system for $\theta$ :

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

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Answer

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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Question

Solve this system for $\theta$ :

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

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Answer

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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Question

Solve this system for $\theta$ :

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

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

Tap to reveal answer

Answer

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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Question

Solve the following system:

Tap to reveal answer

Answer

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.

← Didn't Know|Knew It →

Question

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

where is a real number.

Tap to reveal answer

Answer

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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Question

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

Tap to reveal answer

Answer

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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Question

Solve the system for $\theta$ :

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

Tap to reveal answer

Answer

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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Question

Solve this system for $\theta$ :

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

Tap to reveal answer

Answer

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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Question

Solve this system for $\theta$ :

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

Tap to reveal answer

Answer

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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Question

Solve this system for $\theta$ :

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

Tap to reveal answer

Answer

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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Question

Solve this system for $\theta$ :

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

Tap to reveal answer

Answer

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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Question

Solve the following system:

Tap to reveal answer

Answer

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.

← Didn't Know|Knew It →

Question

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

where is a real number.

Tap to reveal answer

Answer

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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