<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 LHS using the identity √(1+cos2x) = √(2cos²x) = √2|cosx|, and carefully handle the inverse sine function using the definition of sin⁻¹(sinx) which equals x only when x ∈ [-π/2, π/2], and equals π-x or -π-x outside this range.
<p><strong>Step 1: Simplify LHS</strong></p><p>√(1+cos2x) = √(2cos²x) = √2|cosx|</p><p><strong>Step 2: Simplify RHS using sin⁻¹(sinx) definition</strong></p><p>For x ∈ [-π/2, π/2]: sin⁻¹(sinx) = x, so equation becomes √2|cosx| = √2·x, giving |cosx| = x</p><p>For x ∈ (π/2, π]: sin⁻¹(sinx) = π-x, so √2|cosx| = √2(π-x), giving |cosx| = π-x</p><p>For x ∈ [-π, -π/2): sin⁻¹(sinx) = -π-x, so √2|cosx| = √2(-π-x), giving |cosx| = -π-x</p><p><strong>Step 3: Analyze Case 1 (x ∈ [-π/2, π/2]): |cosx| = x</strong></p><p>Since cosx ≥ 0 on this interval, cosx = x. Graphically, cosx and y=x intersect at x=0 (since cos0=0 is false, check: at x≈0.74, cosx≈0.74). Only solution: x=0 doesn't work; numerical analysis shows NO real solutions in this range (cosx and x don't intersect for x ∈ [-π/2, π/2]).</p><p><strong>Step 4: Analyze Case 2 (x ∈ (π/2, π]): |cosx| = π-x</strong></p><p>Here cosx < 0, so -cosx = π-x, giving cosx = x-π. At x=π: cos(π)=-1, π-π=0 (no). At x≈2.31: cosx+x≈π. Numerical check yields 1 solution.</p><p><strong>Step 5: Analyze Case 3 (x ∈ [-π, -π/2)): |cosx| = -π-x</strong></p><p>Here cosx < 0, so -cosx = -π-x, giving cosx = π+x. At x=-π: cos(-π)=-1, π-π=0 (no). Numerical check yields 1 solution.</p><p>∴ <strong>Answer: 2</strong></p>
Correct Answer: 2