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  $\displaystyle x=\frac{5\pi}{6}; \frac{7\pi}{6}$

$\displaystyle 3\tan^2x=1$;  Divide both sides by 3

$\displaystyle \tan^2x=\frac{1}{3}$;  Take the square root on both sides.  Just as the previous question, when you take a square root the answer could be positive or negative.

$\displaystyle \tan x=\pm \frac{1}{\sqrt{3}}$;  This can be written as two separate equations

$\displaystyle \tan x=\frac{1}{\sqrt{3}}$  and  $\displaystyle \tan x=-\frac{1}{\sqrt{3}}$;  Take the inverse tangent

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

$\displaystyle \tan ^2x+2\tan x+1=0$;  The expression is similar to a quadratic expression and can be factored.

$\displaystyle (\tan x+1)(\tan x+1)=0$;  set both expressions equal to 0.  Since they are the same, the solutions will repeat, so I will only write it once.

$\displaystyle \tan x+1=0$

$\displaystyle \tan x=-1$;  take the inverse tangent on both sides

$\displaystyle x=\tan ^{-1}(-1)$

$\displaystyle x=\frac{3\pi}{4}; \frac{7\pi}{4}$

Recall the values of $\displaystyle v$ for which $\displaystyle sin(v)= 0$. If it helps, think of sine as the $\displaystyle y$ values on the unit circle. Thus, the acceptable values of $\displaystyle v$ would be 0, 180, 360, 540 etc.. However, in our scenario $\displaystyle v = 4x$.

Thus we have $\displaystyle 0 = 4x$ and $\displaystyle 180 = 4x$.

Any other answer would give us values greater than 90. When we divide by 4, we get our answers,

$\displaystyle x=0$ and $\displaystyle x=45$. 

$\displaystyle 2\sin x=1$; Divide both sides by 2 to get

$\displaystyle \sin x=\frac{1}{2}$; take the inverse sine on both sides

$\displaystyle \sin^{-1}(\sin x)=\sin ^{-1}\bigg(\frac{1}{2}\bigg)$; the left side reduces to x, so

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

At this point, either use a unit circle diagram or a calculator to find the value.

Keep in mind that the problem asks for all solutions between $\displaystyle 0$ and $\displaystyle 2\pi$.

If you use a calculator, you will only get $\displaystyle \frac{\pi}{6}$ as an answer. 

So we need to find another angle that satisfies the equation $\displaystyle \sin x=\frac{1}{2}$.

$\displaystyle \sin 2x=4\cos x$;  First use the double angle identity for $\displaystyle \sin 2x$.

$\displaystyle 2\sin x \cos x = 4\cos x$;  divide both sides by 2

$\displaystyle \sin x \cos x=2\cos x$;  subtract the $\displaystyle 2\cos x$ from both sides

$\displaystyle \sin x \cos x-2\cos x=0$;  factor out the $\displaystyle \cos x$

$\displaystyle \cos x(\sin x-2)=0$; Now we have the product of two expressions is 0.  This can only happen if one (or both) expressions are equal to 0.  So let each expression equal 0.

$\displaystyle \cos x= 0$  or  $\displaystyle \sin x-2=0$; 

$\displaystyle \cos x = 0$  or  $\displaystyle \sin x=2$;  Take the inverse of each function for each expression.

$\displaystyle x=\cos ^{-1}(0)$ or $\displaystyle x=\sin ^{-1}(2)$;  The second equation is not possible so gives no solution, but the first equation gives us:

$\displaystyle x=\frac{\pi}{2}; \frac{3\pi}{2}$