Solve for $x$; $\cos^2x-\sin^2x=\sin x; -\pi\lt x\leq\pi$ 
Solve for $x$; $\cos^2x-\sin^2x=\sin x; -\pi\lt x\leq\pi$
   $$\cos^2x-\sin^2x=\sin$$

Edit
$$1-\sin^2x-\sin^2x=\sin x$$ 
$$2\sin^2 x+\sin x-1=0$$
$\sin x=a$
$$2a^2+a-1=0$$
$$(a+1)(2a-1)=0$$
$$x=-1,\dfrac{1}{2}$$
$$x=\sin^{-1}(.5)=30^{\circ}=\dfrac{\pi}{6}$$
$$x=sin^{-1}(-1)=-90^{\circ}=-\dfrac{\pi}{2}$$

The calculation is almost completely correct. You reached the two possibilities $\sin x=\frac{1}{2}$ and $\sin x=-1$.
We are interested in solutions in the interval $-\pi \lt x\le \pi$.
Certainly $x=\frac{\pi}{6}$ is a solution, since $\sin(\pi/6)=\frac{1}{2}$. But there is another $x$ in our interval whose sine is $\frac{1}{2}$, namely $x=\pi-\frac{\pi}{6}=\frac{5\pi}{6}$.  A look at the graph of $y=\sin x$ shows this. You can do a partial verification by calculator, by asking it to compute $\sin(5\pi/6)$, the sine of $150^\circ$. 
There is only one place $x$ in our interval where $\sin x=-1$, so that part is fully correct.
