Work shown so far:
This is where I'm stuck at. I know that I somehow have to get two terms to equal to zero, but I don't know how. The +1 at the end of the problem is throwing me for a loop. Am I even doing it correct?
Correct up to $\displaystyle \displaystyle 2 - 2\sin^2{x} + \sin{x} - 1 = 0$.
From here, you have forgotten that after that it is an EQUATION, and also there should be a negative on the $\displaystyle \displaystyle 2\sin^2{x}$.
Solve it as you would a quadratic equation.
By the identity $\displaystyle 1+\cos{2x} = 2\cos^2{x}$, we've $\displaystyle 2\cos^2{x}+\sin{x}-1 = \cos{2x}+\sin{x}$.
So for your equation we have $\displaystyle \cos{2x}+\sin{x} = 0 \Rightarrow \cos{2x} = -\sin{x} = \cos\left(x+\frac{\pi}{2}\right)$.
If $\displaystyle \cos{\theta} = \cos{\alpha}$, then $\displaystyle \theta = 2n\pi\pm \alpha$, thus for the equation $\displaystyle \cos{2x} = \cos\left(x+\frac{\pi}{2}\right)$ we've:
$\displaystyle 2x = 2n\pi+x+\frac{\pi}{2} \Rightarrow x = 2n\pi+\frac{\pi}{2} $ or we've $\displaystyle 2x = 2n\pi-x-\frac{\pi}{2} \Rightarrow x = \frac{2}{3}n\pi-\frac{\pi}{6}.$