How to compute $\sin^{-1}(\cos 2x)$ in different domains? What is $\sin^{-1}(\cos 2x)$ when $x \in [\pi/2,3\pi/2]$?
I tried 
$$\sin^{-1}(\cos 2x)=\sin^{-1}(\sin (\pi/2-2x))=\pi/2-2x$$
However, it turns out to be the solution when $x \in [0, \pi/2]$.
and the solution when $x \in [\pi/2,3\pi/2]$ is 
$$2x-3\pi/2$$
So how to related the solution to different domains?
 A: As $\cos^{-1}y+\sin^{-1}y=\dfrac\pi2$
$\sin^{-1}(\cos2x)=\dfrac\pi2-\cos^{-1}(\cos2x)$
As $\pi\le2x\le3\pi,$  
$\cos^{-1}(\cos2x)=2\pi-2x$ for $\pi\le2x\le2\pi$
For $2\pi<2x\le3\pi,\cos^{-1}(\cos2x)=2x-2\pi$
A: 
Herein, we present a systematic approach to the problem of interest.  To that end we now proceed.

Note that for $\theta\in [(n-1/2)\pi,(n+1/2)\pi]$, we have

$$\bbox[5px,border:2px solid #C0A000]{\arcsin(\sin(\theta))=(-1)^n(\theta-n\pi) }\tag 1$$

Let $\theta=\pi/2-2x$ in $(1)$.  

CASE $1$: $\displaystyle x\in[\pi/2,\pi]$ 

If $x\in [\pi/2,\pi]$, then $\theta\in[-3\pi/2,-\pi/2]$.  Using $(1)$ with $n=-1$ reveals
$$\begin{align}
\arcsin(\cos(2x))&=\arcsin(\sin(\theta))\\\\
&=-(\theta+\pi)\\\\
&=-(\pi/2-2x+\pi)\\\\
&=2x-3\pi/2
\end{align}$$


CASE $2$: $\displaystyle x\in[\pi,3\pi/2]$ 

If $x\in [\pi,3\pi/2]$, then $\theta\in[-5\pi/2,-3\pi/2]$.  Using $(1)$ with $n=-2$ reveals
$$\begin{align}
\arcsin(\cos(2x))&=\arcsin(\sin(\theta))\\\\
&=\theta+2\pi\\\\
&=\pi/2-2x+2\pi\\\\
&=5\pi/2-2x
\end{align}$$
