<p><strong>968.</strong> The number of real solutions of the equation \(\sqrt{1+\cos 2x} = \sqrt{2}\sin^{-1}(\sin x)\) where \(-\pi \leq x \leq \pi\).</p>
Step-by-Step Solution
Key Concept: Simplify the left side using the identity 1 + cos 2x = 2cos²x, then carefully apply the range and properties of inverse sine function sin⁻¹(sin x), which equals x only when x ∈ [-π/2, π/2].
<p><strong>Step 1: Simplify the left side.</strong></p><p>Using the identity 1 + cos 2x = 2cos²x:</p><p>√(1 + cos 2x) = √(2cos²x) = √2|cos x|</p><p><strong>Step 2: Simplify the right side using inverse sine properties.</strong></p><p>For x ∈ [-π, π], we need to determine sin⁻¹(sin x):</p><p>• If x ∈ [-π/2, π/2]: sin⁻¹(sin x) = x</p><p>• If x ∈ (π/2, π]: sin⁻¹(sin x) = π - x</p><p>• If x ∈ [-π, -π/2): sin⁻¹(sin x) = -π - x</p><p><strong>Step 3: Case 1 – When x ∈ [-π/2, π/2].</strong></p><p>Equation becomes: √2|cos x| = √2 · x</p><p>Simplifying: |cos x| = x</p><p>For x ∈ [-π/2, 0]: -cos x = x ⟹ cos x = -x. Solving: x ≈ -0.739 (one solution)</p><p>For x ∈ [0, π/2]: cos x = x. Solving: x ≈ 0.739 (one solution)</p><p><strong>Step 4: Case 2 – When x ∈ (π/2, π].</strong></p><p>Equation becomes: √2|cos x| = √2(π - x)</p><p>Simplifying: |cos x| = π - x</p><p>For x ∈ (π/2, π], cos x ≤ 0, so |cos x| = -cos x:</p><p>-cos x = π - x ⟹ cos x = x - π</p><p>At x = π: cos π = -1 and π - π = 0 (no match)</p><p>At x = π/2: cos(π/2) = 0 and π/2 - π = -π/2 (no match)</p><p>Checking the behavior: π - x ranges from (π/2, 0] as x goes from (π/2, π], but -cos x ranges from (0, 1]. These intervals don't overlap properly for solutions in the interior. No solution in this interval.</p><p><strong>Step 5: Case 3 – When x ∈ [-π, -π/2).</strong></p><p>Equation becomes: √2|cos x| = √2(-π - x)</p><p>Simplifying: |cos x| = -π - x</p><p>For x ∈ [-π, -π/2), we need -π - x ≥ 0, so x ≤ -π. Only x = -π satisfies this constraint.</p><p>At x = -π: |cos(-π)| = |-1| = 1 and -π - (-π) = 0 (no match). No solution.</p><p><strong>∴ Answer: 2</strong></p>
Correct Answer: 2