If 0≤x≤2π and |cos x|≤sinx, then

# If  then

1. A

$x\in \left[0,\frac{\pi }{4}\right]$

2. B

$x\in \left[\frac{\pi }{4},2\pi \right]$

3. C

$\left[\frac{\pi }{4},\frac{3\pi }{4}\right]$

4. D

$\left[0,\pi \right]$

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### Solution:

We have, $|\mathrm{cos}\text{\hspace{0.17em}}x|\le \mathrm{sin}x⇒\mathrm{sin}x\ge 0\left(\because |\mathrm{cos}\text{\hspace{0.17em}}x|\ge 0\right)$

$⇒x\notin \left(\pi ,2\pi \right)$ if $x=2\pi ,|\mathrm{cos}2\pi |\le \mathrm{sin}2\pi$

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

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

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

$⇒-\mathrm{cos}x\le \mathrm{sin}x\text{\hspace{0.17em}\hspace{0.17em}}\left(\because \mathrm{cos}x<0\right)$

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

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

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