Question

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

• A. $x\in [0,\frac{\pi }{4}]$
• B. $x\in [\frac{\pi }{4},\frac{\pi }{2}]$
• C. $x\in [\frac{\pi }{4},\frac{3\pi }{4}]$
• D. None of these

Answer 1

The correct option is C $x\in [\frac{\pi }{4},\frac{3\pi }{4}]$

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

$\therefore x\in [0,\pi ]$

If $x\in [0,\frac{\pi }{2}]$, then

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

If $x\in [\frac{\pi }{2},\pi ]$, then

$\left|\cos x\right|\le \sin x\Rightarrow -\cos x\le \sin x\Rightarrow \tan x\le -1(\because \cos x<0)$

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

Comparing two results, we get

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