Question

The number of real solutions of the equation

$\sqrt{1+\cos 2x}=\sqrt{2}{\cos }^{-1}\left(\cos x\right)$⁡

in$[\frac{\pi }{2},\pi ]$is

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

Answer 1

The correct option is A $0$

We have

$\sqrt{1+\cos 2x}=\sqrt{2}{\cos }^{-1}\left(\cos x\right)$

$\Rightarrow \sqrt{2{\cos }^{2}x}=\sqrt{2}{\cos }^{-1}\left(\cos x\right)$

$\Rightarrow |\cos x|={\cos }^{-1}\left(\cos x\right)$

$\Rightarrow |\cos x|=x$ $[\because {\cos }^{-1}\cos x=x:xϵ[0,\pi \left]\right]$

$\because \cos x\le 0\text{for}x\in [\frac{\pi }{2},\pi ]$

$\Rightarrow \cos x\ne x\text{for}x\in [\frac{\pi }{2},\pi ]$

Hence, there are $0$ real solutions of the given equation.

Therefore, option $\left(a\right)$ is correct.