<p>The number of solutions of the equation \(\sqrt{1 + \cos 2x} = \sqrt{2}\sin^{-1}(\sin x)\) for \(x \in [-\pi, \pi]\) is:</p>
Step-by-Step Solution
Key Concept: Simplify the left side using $\sqrt{1 + \cos 2x} = \sqrt{2}|\cos x|$, and carefully evaluate $\sin^{-1}(\sin x)$ which equals $x$ only when $x \in [-\pi/2, \pi/2]$, but equals $\pi - x$ for $x \in (\pi/2, \pi]$ and $-\pi - x$ for $x \in [-\pi, -\pi/2)$.
<p><strong>Step 1:</strong> Simplify LHS: $\sqrt{1 + \cos 2x} = \sqrt{2\cos^2 x} = \sqrt{2}|\cos x|$</p><p><strong>Step 2:</strong> Equation becomes: $\sqrt{2}|\cos x| = \sqrt{2}\sin^{-1}(\sin x)$, so $|\cos x| = \sin^{-1}(\sin x)$</p><p><strong>Step 3:</strong> Analyze by regions of $x$:</p><p><strong>Case 1:</strong> $x \in [-\pi/2, \pi/2]$: $\sin^{-1}(\sin x) = x$, so $|\cos x| = x$</p><ul><li>For $x \in [0, \pi/2]$: $\cos x = x$ has one solution at $x \approx 0.739$</li><li>For $x \in [-\pi/2, 0]$: $\cos x = -x$ gives $x \approx -0.739$ (one solution)</li></ul><p><strong>Step 4:</strong> $x \in (\pi/2, \pi]$: $\sin^{-1}(\sin x) = \pi - x$, so $|\cos x| = \pi - x$. Since $\cos x < 0$ here, $-\cos x = \pi - x$, giving $\cos x = x - \pi$. Checking: no valid solutions in this range.</p><p><strong>Step 5:</strong> $x \in [-\pi, -\pi/2)$: $\sin^{-1}(\sin x) = -\pi - x$, so $|\cos x| = -\pi - x$. Since $\cos x < 0$ here, $-\cos x = -\pi - x$, giving $\cos x = \pi + x$. Checking: no valid solutions.</p><p>∴ Answer: <strong>B</strong> (2 solutions)</p>
Correct Answer: B