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 $1+\sin(x)=0$
$$\sin(x)=-1 = -90\text{ degrees }  = -\pi/2  \text{ or  }3\pi/2$$

Hey there you are partially correct , one thing you missed in your solution 
Follow my solution 
$\cos^2x=1-\sin x$
$\Longrightarrow (1-\sin x)(1+\sin x)=1-\sin x$
From this equation we have 
$1- \sin x=0$ and $1+ \sin x=0$
$\sin x = 1$ and $\sin x = -1$
$\sin x= \sin(\frac\pi2)$ and $\sin x= \sin(-\frac\pi2)$
We have if $\sin x= \sin\alpha$ then $x= n\pi+(-1)^n\alpha$
Use this result for both equations and find solution.
