Find the values of $x$ satisfying $\sin^{-1}(|\sin x|)-\cos^{-1}(\cos x)\ge0$ in $[0, 2\pi]$ Find the values of $x$ satisfying $\sin^{-1}(|\sin x|)-\cos^{-1}(\cos x)\ge0$ in $[0, 2\pi]$. I think it would be better explained by drawing the graphs.
Kindly help me in this question.
 A: HINT:
First of all, recount the principal values of $\sin^{-x},\cos^{-1}x$
Now, $|\sin x|=+\sin x$ if $0\le x\le\pi$
and $|\sin x|=-\sin x$ if $\pi<x\le2\pi$
Again for  $0\le x\le\pi,\sin^{-1}|\sin x|=x$ if $0\le x\le\dfrac\pi2$ and $=\pi-x$ if $\dfrac\pi2<x\le\pi$
and for $\pi<x\le2\pi,\sin^{-1}|\sin x|=\sin^{-1}(-\sin x)=\sin^{-1}[\sin(-x)]$
Now
$\sin^{-1}[\sin(-x)]=-x$ if $-\dfrac\pi2\le-x\le\dfrac\pi2\iff-\dfrac\pi2\le x\le\dfrac\pi2$ 
and $\pi-x$ if $-\dfrac\pi2\le\pi-x\le\dfrac\pi2\iff\dfrac\pi2\ge x-\pi\ge-\dfrac\pi2\iff\dfrac\pi2\le x\le\dfrac{3\pi}2$
A: For $0\leq x \leq \pi/2$
We can write the above statement simply as 
$x-x \geq 0$
Which is true, 
For next interval, $\pi/2 \leq x \leq \pi$
We can have, 
$\sin x$ is positive in that domain however the $x \notin$ Principal domain. And $\cos x$ is already in it's principal domain. To get it back to principal domain we write it as.
$\sin^{-1}(\sin (\pi - x)- \cos^{-1}(\cos x) $  
=$\pi - x -x$
=$\pi - 2x\leq 0$ $\forall$ values of $x\in [\pi/2 ,\pi]$
I am sure you can take it from here.
A: It's the same if you look at $f(x)=\arcsin(|\sin x|)-\arccos(\cos x)$ in the interval $[-\pi,\pi]$, which also allows for a further simplification, because
$$
f(-x)=\arcsin(|\sin(-x)|)-\arccos(\cos x)=f(x)
$$
Thus we just need to look at the interval $[0,\pi]$ and it seems natural to split the study the two halves.
If $x\in[0,\pi/2]$, we have $|\sin x|=\sin x$ and $\arcsin\sin x=x$; similarly, $\arccos\cos x=x$, so $f(x)=0$.
If $x\in[\pi/2,\pi]$, we have $|\sin x|=\sin x$ and $\arcsin\sin x=\pi-x$; also $\arccos\cos x=x$, so $f(x)=\pi-x-x=\pi-2x\le0$.
