Question

The number of real solutions of $\sqrt{1 - \cos 2x} = \sqrt{2} \sin^{-1}(\sin x), -\pi \le x \le \frac{\pi}{2}$ is

Answer 1

2

Answer 2

The equation $\sqrt{1 - \cos 2x} = \sqrt{2} \sin^{-1}(\sin x)$ is simplified to $\sqrt{2} |\sin x| = \sqrt{2} \sin^{-1}(\sin x)$. This simplifies to $|\sin x| = \sin^{-1}(\sin x)$. The given domain is $-\pi \le x \le \frac{\pi}{2}$. We analyze the equation in two parts of the domain:

1. For $-\pi \le x < -\frac{\pi}{2}$: $|\sin x| = -\sin x$ and $\sin^{-1}(\sin x) = -\pi - x$. The equation becomes $-\sin x = -\pi - x \Rightarrow \sin x = \pi + x$. By analyzing the graphs or functions, $x = -\pi$ is the only solution.

2. For $-\frac{\pi}{2} \le x \le \frac{\pi}{2}$:

   a. For $-\frac{\pi}{2} \le x < 0$: $|\sin x| = -\sin x$ and $\sin^{-1}(\sin x) = x$. The equation becomes $-\sin x = x \Rightarrow \sin x = -x$. This has no solutions in this interval.

   b. For $0 \le x \le \frac{\pi}{2}$: $|\sin x| = \sin x$ and $\sin^{-1}(\sin x) = x$. The equation becomes $\sin x = x$. This equation has only one solution, $x=0$.

Combining all valid solutions, we get $x = -\pi$ and $x = 0$. Therefore, there are 2 real solutions.