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

The given equation is $$\sqrt{1+\cos(2x)}=\sqrt{2}\,\tan^{-1}(\tan x)$$ to be solved for $$x$$ in
$$\left(-\tfrac{3\pi}{2},-\tfrac{\pi}{2}\right)\cup\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)\cup\left(\tfrac{\pi}{2},\tfrac{3\pi}{2}\right).$$

Step 1 : Simplify the left-hand side
Using $$\cos(2x)=2\cos^{2}x-1,$$ we get $$1+\cos(2x)=2\cos^{2}x\; \Longrightarrow\; \sqrt{1+\cos(2x)}=\sqrt{2}\,|\,\cos x|.$$ Hence the equation becomes $$\sqrt{2}\,|\,\cos x|=\sqrt{2}\,\tan^{-1}(\tan x)$$ or after cancelling $$\sqrt{2}$$ $$|\,\cos x|=\tan^{-1}(\tan x). \quad -(1)$$

Step 2 : Evaluate $$\tan^{-1}(\tan x)$$ on the three sub-intervals
Because $$\tan^{-1}(\cdot)$$ returns values only in $$(-\tfrac{\pi}{2},\tfrac{\pi}{2}),$$ we have

Case 1: $$x\in\bigl(-\tfrac{\pi}{2},\tfrac{\pi}{2}\bigr)$$
 $$\tan^{-1}(\tan x)=x.$$ Case 2: $$x\in\bigl(\tfrac{\pi}{2},\tfrac{3\pi}{2}\bigr)$$
 Subtract $$\pi$$ to bring the angle into the principal range:  $$\tan^{-1}(\tan x)=x-\pi.$$ Case 3: $$x\in\bigl(-\tfrac{3\pi}{2},-\tfrac{\pi}{2}\bigr)$$
 Add $$\pi$$ for the same reason:  $$\tan^{-1}(\tan x)=x+\pi.$$

Step 3 : Solve equation (1) in each case

Case 1: $$|\,\cos x|=\cos x\;( \cos x\gt 0)\; \Longrightarrow\; \cos x = x,\quad x\in\bigl(-\tfrac{\pi}{2},\tfrac{\pi}{2}\bigr).$$
Define $$f(x)=\cos x-x.$$ $$f(0)=1\gt0$$ and $$f\!\bigl(\tfrac{\pi}{2}\bigr)=0-\tfrac{\pi}{2}\lt0;$$ $f(x)$$ is continuous and strictly decreasing on this interval, so exactly one root exists. This root lies near $$x$$\approx$$0.739$$ (between 0 and 1).

Case 2: Here $$|\,$$\cos$$ x|=-$$\cos$$ x\;( $$\cos$$ x\lt 0).$$ Equation (1) becomes $$-$$\cos$$ x = x-$$\pi$$,\quad x$$\in$$\bigl(\tfrac{$$\pi$$}{2},\tfrac{3$$\pi$$}{2}\bigr).$$ Put $$y=x-$$\pi$$\;( $$\Rightarrow$$ y$$\in$$(-\tfrac{$$\pi$$}{2},\tfrac{$$\pi$$}{2})).$$ Then $$-$$\cos$$(\,y+$$\pi$$)=y.$$ Because $$$$\cos$$(y+$$\pi$$)=-$$\cos$$ y,$$ this simplifies to $$$$\cos$$ y = y,\quad y$$\in$$\bigl(-\tfrac{$$\pi$$}{2},\tfrac{$$\pi$$}{2}\bigr).$$ This is identical to Case 1 and again has the single root $$y$$\approx$$0.739.$$ Therefore $$x=y+$$\pi$$$$\approx$$0.739+$$\pi$$$$\approx$$3.88,$$ which indeed lies in $$\bigl(\tfrac{$$\pi$$}{2},\tfrac{3$$\pi$$}{2}\bigr).$$

Case 3: Again $$|\,$$\cos$$ x|=-$$\cos$$ x$$ (cosine is negative in this interval). Equation (1) becomes $$-$$\cos$$ x = x+$$\pi$$,\quad x$$\in$$\bigl(-\tfrac{3$$\pi$$}{2},-\tfrac{$$\pi$$}{2}\bigr).$$ Set $$y=x+$$\pi$$\;( $$\Rightarrow$$ y$$\in$$(-\tfrac{$$\pi$$}{2},\tfrac{$$\pi$$}{2})).$$ With $$x=y-$$\pi$$,$$ we have $$-$$\cos$$(y-$$\pi$$)=y.$$ Since $$$$\cos$$(y-$$\pi$$)=-$$\cos$$ y,$$ this again reduces to $$$$\cos$$ y = y,\quad y$$\in$$\bigl(-\tfrac{$$\pi$$}{2},\tfrac{$$\pi$$}{2}\bigr).$$ We get the familiar single root $$y$$\approx$$0.739,$$ giving $$x=y-$$\pi$$$$\approx$$0.739-$$\pi$$$$\approx$$-2.40,$$ which is indeed in $$\bigl(-\tfrac{3$$\pi$$}{2},-\tfrac{$$\pi$$}{2}\bigr).$$

Step 4 : Count the solutions
Each case provides exactly one solution, and the three solutions are distinct: $$x_1$$\approx$$0.739,\; x_2$$\approx$$3.88,\; x_3$$\approx$$-2.40.$$ Hence the total number of real solutions is $$3.$$

Final Answer: 3