Question

The number of real solution of the equation $\sqrt{1+cos2x}=\sqrt{2}si{n}^{-1}\left(sinx\right),-\pi <x\le \pi$, is

• A. $0$
• B. $1$
• C. $2$
• D. $infinite$

Answer 1

The correct option is C $2$

Given equation is,

$\sqrt{1+\cos 2x}=\sqrt{2}si{n}^{-1}\left(\sin x\right)\Rightarrow \sqrt{1+{\cos }^{2}x-{\sin }^{2}x}=\sqrt{2}x\Rightarrow \sqrt{2{\cos }^{2}x}=\sqrt{2}x\Rightarrow \pm \sqrt{2}\cos x=\sqrt{2}x\Rightarrow \cos x=\pm 1x=0,\pi$