Trigonometry - Systems of Trigonometric Equations

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.

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

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

This system does not have a solution.

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

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

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

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

Explanation

We can write the system in the equivalent form:

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

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

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

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.

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

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

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 $-\frac{1}{\sqrt2}$ .

Solve the system for $\theta$ :

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

$\frac{ \pi }{2 }$

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

$2 \pi$

$\pi$

no solution

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

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

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

Solve this system for $\theta$ :

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

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

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

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

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

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

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

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

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

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

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

Solve this system for $\theta$ :

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

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

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

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

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

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

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

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

Solve this system for $\theta$ :

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

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

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

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

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

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

no solution

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

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

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

Solve this system for $\theta$ :

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

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

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

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

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

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

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

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

Solve this system for $\theta$ :

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

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

no solution

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

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

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

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

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

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

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

Solve the following system:

Screen shot 2020 05 21 at 10.30.36 am

The system does not have a solution.

$\frac{\pi}{3}$

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

$\pi$

$-\pi$

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.

Systems of Trigonometric Equations

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