Let ${\tan }^{-1}\left(x\right)\in \left(-\frac{\pi }{2}, \frac{\pi }{2}\right)$, for $x\in \mathrm{ℝ}$. Then the number of real solutions of the equation $\sqrt{1+\cos \left(2x\right)}=\sqrt{2}{\tan }^{-1}\left(\tan x\right)$ in the set $\left(-\frac{3\pi }{2}, -\frac{\pi }{2}\right)\cup \left(-\frac{\pi }{2}, \frac{\pi }{2}\right)\cup \left(\frac{\pi }{2}, \frac{3\pi }{2}\right)$ is equal to

Given,

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

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

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

Now plotting the diagram of $\left|\cos x\right| \& {\tan }^{-1}\left(\tan x\right)$ we get,

Now from above diagram we can say that,

Number of solutions $=$ Number of intersection points $=3$