If $0\le x\le 2\pi and |\cos x|\le \sin x,$ then

Solution:

We have, $\left|\cos x\right|\le \sin x\Rightarrow \sin x\ge 0(\because \left|\cos x\right|\ge 0)$

$\Rightarrow x\notin (\pi ,2\pi )$ if $x=2\pi ,\left|\cos 2\pi \right|\le \sin 2\pi$

Which is not possible $\therefore x\in \left[0,\frac{\pi }{2}\right],then$

$\left|\cos x\right|\le \sin x\Rightarrow x\in \left[\frac{{\displaystyle \pi }}{{\displaystyle 4}},\frac{{\displaystyle \pi }}{{\displaystyle 2}}\right]$

If $x\in \left(\frac{{\displaystyle \pi }}{{\displaystyle 2}},\pi \right),$ then $\left|\cos x\right|\le \sin x$

$\Rightarrow -\cos x\le \sin x(\because \cos x<0)$

$x\in (\frac{\pi }{2},\frac{3\pi }{4}]$

$\Rightarrow x\in [\frac{\pi }{4},\frac{\pi }{2}]\cup (\frac{\pi }{2},\frac{3\pi }{4}]$