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 $1+\sin(x)=0$
$$\sin(x)=-1 = -90\text{ degrees }  = -\pi/2  \text{ or  }3\pi/2$$
 A: You are almost correct: at the step: $1 - \sin^2 x = 1 - \sin x$ It would be easier to do the following: $$\sin^2 x - \sin x = \sin{x} (\sin{x} - 1) = 0$$
Thus, we would have $\sin{x} = 0$ and $\sin{x} = 1$ 
From here it follows that your solutions are, in the domain $[0, 2\pi)$, 
$$0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}$$
Unfortunately, you potentially divided by 0 without accounting for it: you have to consider when $\sin{x} = 1$; Also, pay mind to your domain. 
A: If you directly divide by $1-\sin x$, you miss the solution $x=\frac{\pi}{2}$, since it makes both side zero. After noting this, you should say $1+\sin x=1$ not $0$. This gives $\sin x=0$, which gives you another solution $x=\pi$.
A: When you divide by $1-\sin x$ you should get $1+\sin x = 1$, not $0$.
A: 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
$$(1-\sin x)(1+\sin x) = 1-\sin(x)$$
$$(1-\sin x)(1+\sin x) - (1-\sin x) = 0$$
$$(1-\sin x)(1+\sin x - 1) = 0$$
$$(1-\sin x)\sin x = 0$$
$$1 - \sin x = 0 \; \text {or} \; \sin x = 0\ldots$$
A: 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.
