The number of real solutions of the equation `sqrt(1+cos2x)=sqrt(2)sin^(-1)(sinx),-pilt=xlt=pi` is 0 (b) 1 (c) 2 (d) infinite

Explanation

To solve the equation \( \sqrt{1 + \cos 2x} = \sqrt{2} \sin^{-1}(\sin x) \) for the number of real solutions in the interval \( -\pi < x < \pi \), we will follow these steps:

### Step 1: Simplify the Left Side
We start with the left side of the equation:
\[ \sqrt{1 + \cos 2x} \]
Using the identity \( \cos 2x = 2\cos^2 x - 1 \), we can rewrite it as:
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