Let tan⁻¹(x) ∈ (−π/2, π/2), for x ∈ ℝ. Then the number of real solutions of the equation

√(1 + cos(2x)) = √2 tan⁻¹(tan x)

in the set (−3π/2, −π/2) ∪ (−π/2, π/2) ∪ (π/2, 3π/2) is equal to

The correct answer is 3.

Let us analyze the given equation step-by-step for real solutions in the specified domain.

The given equation is:

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

Step 1: Simplify the Left-Hand Side (LHS)

Using the trigonometric identity $1+\cos \left(2x\right)=2{\cos }^{2}x$, we can rewrite the left-hand side as:

$\text{LHS}=\sqrt{2{\cos }^{2}x}=\sqrt{2}\left|\cos x\right|$

Step 2: Simplify the Right-Hand Side (RHS)

The right-hand side is given by:

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

Equating LHS and RHS and dividing both sides by $\sqrt{2}$, the equation simplifies to:

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

Step 3: Analyze in each interval of the domain

The given domain is $x\in (-3\pi /2,-\pi /2)\cup (-\pi /2,\pi /2)\cup (\pi /2,3\pi /2)$.

Note that ${\tan }^{-1}\left(\tan x\right)$ is a periodic function with period $\pi$, and its principal range is $(-\pi /2,\pi /2)$. Also, $\left|\cos x\right|\ge 0$, so solutions can only exist where ${\tan }^{-1}\left(\tan x\right)\ge 0$.

Case 1: Interval $x\in (-\pi /2,\pi /2)$

In this interval, ${\tan }^{-1}\left(\tan x\right)=x$ and $\cos x>0$, so $\left|\cos x\right|=\cos x$.

The equation becomes:

$\cos x=x$

Since $y=\cos x$ is strictly decreasing from 1 to 0 on $[0,\pi /2)$ while $y=x$ increases from 0 to $\pi /2$, there is exactly 1 solution in $(0,\pi /2)$. (For $x<0$, $\cos x>0$ but $x<0$, so no negative solutions exist here).

Case 2: Interval $x\in (\pi /2,3\pi /2)$

In this interval, ${\tan }^{-1}\left(\tan x\right)=x-\pi$.

The equation becomes:

$\left|\cos x\right|=x-\pi$

Let $t=x-\pi$. As $x\in (\pi /2,3\pi /2)$, we have $t\in (-\pi /2,\pi /2)$.

Also, $\left|\cos x\right|=\left|\cos \right(t+\pi \left)\right|=|-\cos t|=\left|\cos t\right|=\cos t$ since $t\in (-\pi /2,\pi /2)$.

So the equation becomes $\cos t=t$ for $t\in (-\pi /2,\pi /2)$.

This gives exactly 1 solution for $t$ in $(0,\pi /2)$, which corresponds to 1 solution for $x$ in $(\pi ,3\pi /2)$.

Case 3: Interval $x\in (-3\pi /2,-\pi /2)$

In this interval, ${\tan }^{-1}\left(\tan x\right)=x+\pi$.

The equation becomes:

$\left|\cos x\right|=x+\pi$

Let $u=x+\pi$. As $x\in (-3\pi /2,-\pi /2)$, we have $u\in (-\pi /2,\pi /2)$.

Similarly, $\left|\cos x\right|=\left|\cos \right(u-\pi \left)\right|=\cos u$.

So the equation becomes $\cos u=u$ for $u\in (-\pi /2,\pi /2)$.

This gives exactly 1 solution for $u$ in $(0,\pi /2)$, which corresponds to 1 solution for $x$ in $(-\pi ,-\pi /2)$.

Conclusion:

Adding the solutions from all three sub-intervals gives:

$\text{Total number of real solutions}=1+1+1=3$