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

Let tan⁻¹(x) ∈ (-π/2, π/2), for x ∈ R. Then the number of real solutions of the equation √1+cos(2x) = √2tan⁻¹(tan x) in the set (-3π/2, -π/2) ∪ (-π/2, π/2) ∪ (π/2, 3π/2) is equal to $\tan^{-1}x \in \left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right),\quad x\in\mathbb{R}$ $\sqrt{1+\cos 2x} = \sqrt2\tan^{-1}(\tan x)$ $x \in \left(-\dfrac{3\pi}{2},-\dfrac{\pi}{2}\right)\cup\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)\cup\left(\dfrac{\pi}{2},\dfrac{3\pi}{2}\right)$ $\text{To find: number of real solutions}$ $1.\ \sqrt{1+\cos 2x} = \sqrt2|\cos x|$ $2.\ |\cos x| = \tan^{-1}(\tan x)$ $3.\ \text{solve interval-wise, count}$ $1+\cos 2x = 2\cos^2 x \Rightarrow \sqrt{1+\cos 2x} = \sqrt2\,|\cos x|$ $|\cos x| = \tan^{-1}(\tan x)$ $\tan^{-1}(\tan x) = x - k\pi \text{ for } x\in\left(k\pi-\frac\pi2,\ k\pi+\frac\pi2\right)$ $u = x - k\pi \in \left(-\frac\pi2,\frac\pi2\right):\quad \cos u = u$ $f(u) = \cos u - u,\quad f'(u) = -\sin u - 1 < 0$ $f(0) = 1 > 0,\ f\!\left(\tfrac\pi2\right) = -\tfrac\pi2 < 0 \Rightarrow \text{one root}$ $x = u^*,\quad u^*+\pi,\quad u^*-\pi \qquad (u^* \approx 0.739)$ $\boxed{3}$