The number of real solutions of the equation √(1+cos 2x) = √2 cos⁻¹(cos x) in [π/2, π] is:

The number of real solutions of the equation √(1+cos 2x) = √2 cos⁻¹(cos x) in [π/2, π] is:

• A. 0
• B. 1
• C. 2
• D. 3

Answer: B) 1

Explanation: LHS = √(2cos²x) = √2|cos x| = −√2 cos x (since cos x ≤ 0 in [π/2, π]). RHS = √2 cos⁻¹(cos x). For x ∈ [π/2, π], cos⁻¹(cos x) = 2π−x? No, in [0,π], cos⁻¹(cos x)=x. So RHS = √2 x. Equation: −√2 cos x = √2 x → cos x = −x. Drawing graphs of y=cos x and y=−x in [π/2, π] shows 1 intersection.