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

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

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

\(\displaystyle -1\le sinx\le 1\) implies that :

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

\(\displaystyle -1\le cos^4x+sin^3x\le 2\)

This means that max=2 and min=-1

We can write the system in the equivalent form:

\(\displaystyle cosx\le \frac{\sqrt{2}}{2}\)

\(\displaystyle sinx\le \frac{\sqrt{2}}{2}\)

\(\displaystyle cosx< sinx\)

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

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

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

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

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

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

First, set both equations equal to each other:

\(\displaystyle 6 \sin \theta - 1 = \sin ^2 \theta + 4\) subtract \(\displaystyle 6 \sin \theta\) from both sides

\(\displaystyle -1 = \sin ^2 \theta - 6 \sin \theta + 4\) add 1 to both sides

\(\displaystyle 0 = \sin ^2 \theta - 6 \sin \theta + 5\)

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

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

This gives us 2 potential solutions for \(\displaystyle \sin \theta\):

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

\(\displaystyle \sin \theta = \frac{2}{2 } = 1\)

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

First, set the two equations equal to each other

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

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

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

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

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

\(\displaystyle \theta = \frac{ 11 \pi }{3} , \frac{ 7 \pi }{3}\)

Set the two equations equal to each other

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

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

\(\displaystyle \sin \theta = \pm \frac{ 1}{ \sqrt2}\)

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

Set both equations equal to each other:

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

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

\(\displaystyle \cos ^2 \theta - \sin ^2 \theta = - \frac{1}{2}\)

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

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

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

\(\displaystyle 2 \theta = \frac{ 2 \pi }{3} , \frac{4 \pi }{3 }\) divide by 2

\(\displaystyle \theta = \frac{\pi }{3} , \frac{ 2 \pi }{3}\)

Set the two equations equal to each other:

\(\displaystyle 4 \cos ^2 \theta + \cos \theta - 5 = \cos \theta - 2\) subtract \(\displaystyle \cos \theta\) from both sides

\(\displaystyle 4 \cos ^ 2 \theta - 5 = -2\) add 5 to both sides

\(\displaystyle 4 \cos ^ 2 \theta = 3\) divide both sides by 4

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

\(\displaystyle \cos \theta = \sqrt{\frac{ 3}{4} } = \frac{ \sqrt{3} }{\sqrt{4}} = \pm \frac{ \sqrt{3}}{2}\)

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

\(\displaystyle \theta = \frac{ \pi }{6} , \frac{ 11 \pi }{6}\)

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

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

\(\displaystyle cosx\ge \frac{4}{3}\)

We know that \(\displaystyle cosx\le 1\) 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.

First, set both equations equal to each other:

\(\displaystyle \sin ^2 \theta + 1 = \cos \theta + 2\) subtract \(\displaystyle \cos \theta\) from both sides

\(\displaystyle \sin ^2 \theta - \cos \theta + 1 = 2\)
 

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

\(\displaystyle 1 - \cos^2 \theta - \cos \theta + 1 =2\) combine like terms

\(\displaystyle - \cos ^2 \theta - \cos \theta + 2 = 2\) subtract 2 from both sides

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

We can solve for \(\displaystyle \cos \theta\) using the quadratic formula:

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

This gives us 2 possible values for cosine

\(\displaystyle \cos \theta = \frac{ 1 + 1 }{ -2 } = \frac{ 2 }{ - 2 } = -1\)

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

\(\displaystyle \cos \theta = \frac{ 1 - 1 }{-2 } = \frac{ 0 }{-2} = 0\)

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