Number of Real Solutions to Inverse Trigonometric Equation

To find the number of real solutions to the equation $\sqrt{1+\cos(2x)} = \sqrt{2} \tan^{-1}(\tan x)$ in the domain $S = \left(-\frac{3\pi}{2}, -\frac{\pi}{2}\right) \cup \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \cup \left(\frac{\pi}{2}, \frac{3\pi}{2}\right)$, we first simplify the left-hand side using the trigonometric identity $1 + \cos(2x) = 2\cos^2 x$:

$\sqrt{1+\cos(2x)} = \sqrt{2\cos^2 x} = \sqrt{2}|\cos x|$

Substituting this back into the equation yields: $\sqrt{2}|\cos x| = \sqrt{2}\tan^{-1}(\tan x)$

Since $\sqrt{2} \neq 0$, we can divide both sides by $\sqrt{2}$: $|\cos x| = \tan^{-1}(\tan x)$

Now, we analyze the equation in each of the three disjoint open intervals that form $S$.

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Interval 1: $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

In this interval, $\tan^{-1}(\tan x) = x$. The equation becomes: $|\cos x| = x$

Since the left-hand side $|\cos x| \ge 0$, any solution must satisfy $x \ge 0$. For $x \in \left[0, \frac{\pi}{2}\right)$, $\cos x \ge 0$, so $|\cos x| = \cos x$.

Thus, the equation reduces to: $\cos x = x$

Consider the function $f(x) = \cos x - x$ on $\left[0, \frac{\pi}{2}\right)$:

• $f(0) = 1 > 0$
• $f\left(\frac{\pi}{2}\right) = -\frac{\pi}{2} < 0$
• $f'(x) = -\sin x - 1 < 0$ for all $x \in \left(0, \frac{\pi}{2}\right)$, meaning $f(x)$ is strictly decreasing.

By the Intermediate Value Theorem and monotonicity, $f(x) = 0$ has exactly $1$ real solution in $\left[0, \frac{\pi}{2}\right)$. Let this unique solution be $\alpha \approx 0.739$.

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Interval 2: $x \in \left(\frac{\pi}{2}, \frac{3\pi}{2}\right)$

In this interval, $\tan^{-1}(\tan x) = x - \pi$. The equation becomes: $|\cos x| = x - \pi$

Let $y = x - \pi$. As $x \in \left(\frac{\pi}{2}, \frac{3\pi}{2}\right)$, we have $y \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Substituting $x = y + \pi$ gives: $|\cos(y + \pi)| = y \implies |-\cos y| = y \implies |\cos y| = y$

This equation is identical to the one solved in Interval 1. It has a unique solution $y = \alpha \in \left[0, \frac{\pi}{2}\right)$. Correspondingly, $x = \pi + \alpha \in \left[\pi, \frac{3\pi}{2}\right) \subset \left(\frac{\pi}{2}, \frac{3\pi}{2}\right)$.

Thus, there is exactly $1$ real solution in this interval.

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Interval 3: $x \in \left(-\frac{3\pi}{2}, -\frac{\pi}{2}\right)$

In this interval, $\tan^{-1}(\tan x) = x + \pi$. The equation becomes: $|\cos x| = x + \pi$

Let $y = x + \pi$. As $x \in \left(-\frac{3\pi}{2}, -\frac{\pi}{2}\right)$, we have $y \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Substituting $x = y - \pi$ gives: $|\cos(y - \pi)| = y \implies |-\cos y| = y \implies |\cos y| = y$

Again, this reduces to the same equation, which has a unique solution $y = \alpha \in \left[0, \frac{\pi}{2}\right)$. Correspondingly, $x = -\pi + \alpha \in \left[-\pi, -\frac{\pi}{2}\right) \subset \left(-\frac{3\pi}{2}, -\frac{\pi}{2}\right)$.

Thus, there is exactly $1$ real solution in this interval.

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Conclusion

Summing the solutions from all three intervals, the total number of real solutions is: $1 + 1 + 1 = 3$