Question

If $\cos x=\sqrt{1-\sin 2x},0\le x\le \pi$, then possible value of $x$ is

• A. $\pi$
• B. $0$
• C. ${\tan }^{-1}2$
• D. $3\pi$

Answer 1

Answer: (B), (C)

${\tan }^{-1}2$
$0$

$\cos x=\sqrt{1-\sin \left(2x\right)}$ $x\in [0,x]$

$=\sqrt{1-2\sin x-\cos x}$ $\{\because 1={\sin }^{2}x+{\cos }^{2}x\forall x\in R\}$

$=\sqrt{{\sin }^{2}x-2\sin x-\cos x+{\cos }^{2}x}$

$=\sqrt{[\sin x-\cos x{]}^2}$ $\{\because \sqrt{x}=|x|\}$

$\cos x=|\sin x-\cos x|$

Case I : ${}_1\sin x\ge \cos x:x\in [0,x]$

$\Rightarrow |\sin x-\cos x|=\sin x-\cos x$

$\therefore \cos x=\sin x-\cos x$

$\Rightarrow 2\cos x=\sin x\iff \tan x=2$

$\therefore x={\tan }^{-1}\left(2\right)$ $[\because x\in [0,x\left]\right]$

Case III:

$\sin x<\cos x;x\in [0,x]$

$\Rightarrow |\sin x-\cos x|=\cos x-\sin x$

$\therefore \cos x=\cos x-\sin x$

$\therefore \sin x=0$

$\Rightarrow x=0,\pi$

But $x=\pi$ is rejected as $\cos \left(x\right)=-1$ and $0\ne -1$

$\therefore$ only $x=0$

Finally : $x={\tan }^{-1}\left(2\right),0$