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

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

We begin by substituting a new variable \(\displaystyle u=2x\).

\(\displaystyle \cos 2u=\cos u\);  Use the double angle identity for \(\displaystyle \cos 2u\).

\(\displaystyle 2\cos ^2 u-1=\cos u\);  subtract the \(\displaystyle \cos u\) from both sides.

\(\displaystyle 2\cos ^2 u-\cos u-1=0\);  This expression can be factored.

\(\displaystyle (2\cos u+1)(\cos u-1)=0\);  set each expression equal to 0.

\(\displaystyle 2\cos u+1=0\) or \(\displaystyle \cos u -1=0\);  solve each equation for \(\displaystyle \cos u\)

\(\displaystyle \cos u = -\frac{1}{2}\) or \(\displaystyle \cos u = 1\);  Since we sustituted a new variable we can see that if \(\displaystyle 0\leq x< 2\pi\), then we must have \(\displaystyle 0\leq 2x < 4\pi\).  Since \(\displaystyle u=2x\), that means \(\displaystyle 0\leq u< 4\pi\).

This is important information because it tells us that when we solve both equations for u, our answers can go all the way up to \(\displaystyle 4\pi\) not just \(\displaystyle 2\pi\).

So we get

\(\displaystyle 2x = u=0; \frac{2\pi}{3}; \frac{4\pi}{3}; 2\pi; \frac{8\pi}{3}; \frac{10\pi}{3};\)  Divide everything by 2 to get our final solutions

\(\displaystyle x=0; \frac{\pi}{3}; \frac{2\pi}{3}; \pi; \frac{4\pi}{3}; \frac{5\pi}{3};\)

\(\displaystyle \sin\left(\frac{x}{2}\right)=\cos x -1\);  We start by substituting a new variable.  Let \(\displaystyle u=\frac{x}{2}\).

\(\displaystyle \sin u=\cos 2u-1\); Use the double angle identity for cosine

\(\displaystyle \sin u = 1-2\sin ^2 u-1\);  the 1's cancel, so add \(\displaystyle 2\sin ^2u\) to both sides

\(\displaystyle 2\sin ^2u+\sin u=0\);  factor out a \(\displaystyle \sin u\) from both terms.

\(\displaystyle \sin u(2\sin u+1)=0\);  set each expression equal to 0.

\(\displaystyle \sin u =0\)  or  \(\displaystyle 2\sin u +1=0\);  solve the second equation for sin u.

\(\displaystyle \sin u = 0\)  or  \(\displaystyle \sin u = -\frac{1}{2}\);  take the inverse sine to solve for u (use a unit circle diagram or a calculator)

\(\displaystyle \frac{x}{2}=u=0, \frac{7\pi}{6}, \frac{11\pi}{6}\);  multiply everything by 2 to solve for x.

\(\displaystyle x=0, \frac{7\pi}{3}, \frac{11\pi}{3}\);  Notice that the last two solutions are not within our range  \(\displaystyle 0\leq x< 2\pi\).  So the only solution is \(\displaystyle x=0\).

\(\displaystyle 4\cos ^2x=3\);  First divide both sides of the equation by 4

\(\displaystyle \cos ^2x=\frac{3}{4}\);  Next take the square root on both sides.  Be careful. Remember that when YOU take a square root to solve an equation, the answer could be positive or negative.  (If the square root was already a part of the equation, it usually only requires the positive square root.  For example, the solutions to \(\displaystyle x^2=4\) are 2 and -2, but if we plug in 4 into the function \(\displaystyle f(x)=\sqrt{x}\) the answer is only 2.)  So,

\(\displaystyle \cos x = \pm \frac{\sqrt{3}}{2}\);  we can separate this into two equations

\(\displaystyle \cos x = \frac{\sqrt{3}}{2}\)  and  \(\displaystyle \cos x =- \frac{\sqrt{3}}{2}\);  we get

\(\displaystyle x=\frac{\pi}{6}; \frac{11\pi}{6}\) and  \(\displaystyle x=\frac{5\pi}{6}; \frac{7\pi}{6}\)