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

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

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

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

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

We begin by getting the right side of the equation to equal zero.

\(\displaystyle 2sin^2(x)-sinx-1=0\)

Next we factor.

\(\displaystyle (2sin(x)+1)(sin(x)-1)=0\)

We then set each factor equal to zero and solve.

\(\displaystyle 2sin(x)+1=0\)            or        \(\displaystyle sin(x)-1=0\)

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

We then determine the angles that satisfy each solution within one revolution.

The angles \(\displaystyle \frac{7\pi}{6}\) and \(\displaystyle \frac{11\pi}{6}\) satisfy the first, and \(\displaystyle \frac{\pi}{2}\) satisfies the second.  Only \(\displaystyle \frac{11\pi}{6}\) is among our answer choices.

The fastest way to solve this problem is to substitute a new variable.  Let \(\displaystyle u=2x\).

The equation now becomes:

\(\displaystyle \sin u=\cos u\)

So at what angles are the sine and cosine functions equal.  This occurs at

\(\displaystyle u=\frac{\pi}{4}; \frac{5\pi}{4}; \frac{9\pi}{4}; \frac{13\pi}{4}\)

You may be wondering, "Why did you include

\(\displaystyle \frac{9\pi}{4}; \frac{13\pi}{4}\) if they're not between \(\displaystyle 0\) and \(\displaystyle 2\pi\)?"

The reason is because once we substitute back the original variable, we will have to divide by 2.  This dividing by 2 will bring the last two answers within our range.

\(\displaystyle 2x=\frac{\pi}{4}; 2x=\frac{5\pi}{4}; 2x=\frac{9\pi}{4}; 2x=\frac{13\pi}{4}\)

Dividing each answer by 2 gives us

\(\displaystyle x=\frac{\pi}{8}; \frac{5\pi}{8}; \frac{9\pi}{8}; \frac{13\pi}{8}\)