1. Simplify the left-hand side

Given $\sqrt{1+\cos 2x} = \sqrt{2}\,\tan^{-1}(\tan x).$

Use $1+\cos 2x = 2\cos^2 x.$ So, $\sqrt{1+\cos 2x} = \sqrt{2\cos^2 x} = \sqrt{2}\,|\cos x|.$

Hence the equation becomes $\sqrt{2}\,|\cos x| = \sqrt{2}\,\tan^{-1}(\tan x).$ Dividing by $\sqrt{2}$, $|\cos x| = \tan^{-1}(\tan x).$

2. Understand $\tan^{-1}(\tan x)$

Since $\tan^{-1}(y) \in \left(-\frac\pi2,\frac\pi2\right)$, the value of $\tan^{-1}(\tan x)$ is the unique angle coterminal with $x$ lying in that interval.

So on the given three intervals:

• For $x \in \left(-\frac\pi2,\frac\pi2\right)$, $\tan^{-1}(\tan x)=x.$

• For $x \in \left(\frac\pi2,\frac{3\pi}2\right)$, $\tan^{-1}(\tan x)=x-\pi.$

• For $x \in \left(-\frac{3\pi}2,-\frac\pi2\right)$, $\tan^{-1}(\tan x)=x+\pi.$

Thus we solve piecewise.

3. Case 1: $x \in \left(-\frac\pi2,\frac\pi2\right)$

Equation becomes $|\cos x| = x.$ But on this interval, $\cos x \ge 0$, so $|\cos x|=\cos x$. Hence $\cos x = x.$

Now $\cos x \ge 0$ while RHS $x$ must also be nonnegative, so $x\in[0,\frac\pi2)$. Consider $f(x)=\cos x - x.$ Then $f'(x)=-\sin x-1<0$ for $x\in[0,\frac\pi2)$, so $f$ is strictly decreasing. Also, $f(0)=1>0, \qquad f\left(\frac\pi2\right)= -\frac\pi2<0.$ Therefore there is exactly one solution in this interval.

4. Case 2: $x \in \left(\frac\pi2,\frac{3\pi}2\right)$

Equation becomes $|\cos x| = x-\pi.$ Since RHS must be nonnegative, $x-\pi \ge 0 \implies x\in[\pi,\tfrac{3\pi}2).$ On $[\pi,\tfrac{3\pi}2)$, $\cos x\le 0$, so $|\cos x|=-\cos x.$ Thus $-\cos x = x-\pi.$ Let $u=x-\pi,$ so $u\in[0,\tfrac\pi2)$ and $-\cos x=-\cos(u+\pi)=\cos u.$ Hence the equation becomes $\cos u = u.$ As in Case 1, this has exactly one solution.

So this interval contributes one solution.

5. Case 3: $x \in \left(-\frac{3\pi}2,-\frac\pi2\right)$

Equation becomes $|\cos x| = x+\pi.$ Again RHS must be nonnegative, so $x+\pi \ge 0 \implies x\in[-\pi,-\tfrac\pi2).$ On $[-\pi,-\tfrac\pi2)$, $\cos x\le 0$, hence $|\cos x|=-\cos x.$ So $-\cos x = x+\pi.$ Let $v=x+\pi,$ so $v\in[0,\tfrac\pi2)$ and $x=v-\pi$. Then $-\cos x = -\cos(v-\pi)=\cos v.$ Thus the equation becomes $\cos v = v.$ Again this has exactly one solution.

So this interval also contributes one solution.

6. Total number of real solutions

Adding all three cases: $1+1+1=3.$

Therefore, the number of real solutions is $\boxed{3}.$

7. Comparison with stored answer

Stored correct answer = $3$.

Our derived answer also equals $3$, so they agree.