Can someone please check my work?: $\cos^2(x)=1-\sin(x)$ $$\begin{align}\cos^2(x)&=1-\sin(x)\\
1-\sin^2(x)&=1-\sin(x)\\
(1-\sin x)(1+\sin x)&= 1-\sin(x)
\end{align}$$
divide both sides by $1 - \sin(x)$
End up with $1 + \sin(x)$
The answer is supposed to be in radians between $0$ and $2 \pi$.
So I get $1+\sin(x)=0$
$$\sin(x)=-1 = -90\text{ degrees }  = -\pi/2  \text{ or  }3\pi/2$$

There are two correct ways to proceed from 
$$(1-\sin x)(1+\sin x) = 1-\sin(x)$$
first way
This will be true if 
$1 - \sin x = 0$
$x \in \{90^{\circ}, 270^{\circ}\}$
If not, then we can divide both sides by $1 - \sin x$, getting
$1 + \sin x = 1$
$\sin x = 0$
$x \in \{0^{\circ}, 180^{\circ}\}$
second way
$$(1-\sin x)(1+\sin x) = 1-\sin(x)$$
$$(1-\sin x)(1+\sin x) - (1-\sin x) = 0$$
$$(1-\sin x)(1+\sin x - 1) = 0$$
$$(1-\sin x)\sin x = 0$$
$$1 - \sin x = 0 \; \text {or} \; \sin x = 0\ldots$$
