Let $\tan^{-1}(x)\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$, 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 $\left(-\dfrac{3\pi}{2},-\dfrac{\pi}{2}\right)\cup\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)\cup\left(\dfrac{\pi}{2},\dfrac{3\pi}{2}\right)$
is equal to ___.
Step-by-Step Solution
Key Concept: tan⁻¹(tan x) = x − nπ on each branch; equation reduces to cos u = u on each valid sub-interval
$\sqrt{1+\cos 2x}=\sqrt{2|\cos x|}$. So equation becomes $|\cos x|=\tan^{-1}(\tan x)$.
On each interval $(n\pi-\pi/2,\,n\pi+\pi/2)$, $\tan^{-1}(\tan x)=x-n\pi$.
**Interval $(-\pi/2,\pi/2)$** ($n=0$): $|\cos x|=x$, i.e., $\cos x=x$ (since $\cos x>0$ and $x$ is positive at solution). One solution $x\approx0.739$. ✓
**Interval $(\pi/2,3\pi/2)$** ($n=1$): $|\cos x|=x-\pi$.
- On $(\pi/2,\pi)$: $x-\pi<0\leq|\cos x|$. No solution.
- On $(\pi,3\pi/2)$: let $u=x-\pi\in(0,\pi/2)$. $|\cos(u+\pi)|=|-\cos u|=\cos u=u$. One solution. ✓
**Interval $(-3\pi/2,-\pi/2)$** ($n=-1$): $|\cos x|=x+\pi$.
- On $(-3\pi/2,-\pi)$: $x+\pi<0$. No solution.
- On $(-\pi,-\pi/2)$: $x+\pi\in(0,\pi/2)>0$. $|\cos x|=-\cos x$ (since $\cos x<0$ here). $-\cos x=x+\pi$. Let $v=-(x+\pi)\in(0,\pi/2)$: $\cos(-\pi-v+\pi)... \cos(x)=\cos(-\pi-v)=\cos(\pi+v)=-\cos v$. So $\cos v=x+\pi=-v$? No: $-\cos(x)=x+\pi$. With $x=-\pi+w$, $w\in(0,\pi/2)$: $-\cos(-\pi+w)=w\Rightarrow\cos w=w$. One solution. ✓
Total: **3** solutions.
Correct Answer: 3