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?
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 $2\pi<2x\le3\pi,\cos^{-1}(\cos2x)=2x-2\pi$
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
$$\begin{align} \arcsin(\cos(2x))&=\arcsin(\sin(\theta))\\\\ &=\theta+2\pi\\\\ &=\pi/2-2x+2\pi\\\\ &=5\pi/2-2x \end{align}$$
