# The number of real solutions of the equation

Question:

The number of real solutions of the equation

$\sqrt{1+\cos 2 x}=\sqrt{2} \sin ^{-1}(\sin x),-\pi \leq x \leq \pi$ is

(a) 0
(b) 1
(c) 2
(d) infinite

Solution:

(c) 2

For, $-\pi \leq x \leq \frac{-\pi}{2}$

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

$\Rightarrow \sqrt{2}|\cos x|=\sqrt{2}(-\pi-\mathrm{x})$

$\Rightarrow \sqrt{2}(-\cos x)=\sqrt{2}(-\pi-\mathrm{x})$

$\Rightarrow \cos x=\pi+\mathrm{x}$

It does not satisfy for any value of $x$ in the interval $\left(-\pi, \frac{-\pi}{2}\right)$

For, $\frac{-\pi}{2} \leq x \leq \frac{\pi}{2}$

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

$\Rightarrow \sqrt{2}|\cos x|=\sqrt{2}(\mathrm{x})$

$\Rightarrow \sqrt{2}(\cos x)=\sqrt{2}(\mathrm{x})$

$\Rightarrow \cos x=\mathrm{x}$

It gives one value of $x$ in the interval $\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$

For, $\frac{\pi}{2} \leq x \leq \pi$

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

$\Rightarrow \sqrt{2}|\cos x|=\sqrt{2}(-\pi-\mathrm{x})$

$\Rightarrow \sqrt{2}(-\cos x)=\sqrt{2}(\pi-\mathrm{x})$

$\Rightarrow \cos x=-\pi+\mathrm{x}$

It gives one value of $x$ in the interval $\left(\frac{\pi}{2}, \pi\right)$

$\therefore \sqrt{1+\cos 2 x}=\sqrt{2} \sin ^{-1}(\sin x)$ gives two real solutions in the interval $[-\pi, \pi]$