Question
Find the number of real solutions of:
\[ \sqrt{1+\cos 2x} = \sqrt{2}\,\sin^{-1}(\sin x), \quad -\pi \le x \le \pi \]
Solution
Simplify LHS:
\[ 1 + \cos 2x = 2\cos^2 x \]
\[ \sqrt{1+\cos 2x} = \sqrt{2\cos^2 x} = \sqrt{2}\,|\cos x| \]
So equation becomes:
\[ \sqrt{2}|\cos x| = \sqrt{2}\,\sin^{-1}(\sin x) \]
\[ |\cos x| = \sin^{-1}(\sin x) \]
Now use principal value:
\[ \sin^{-1}(\sin x) = \begin{cases} x, & -\frac{\pi}{2} \le x \le \frac{\pi}{2} \\ \pi – x, & \frac{\pi}{2} < x \le \pi \\ -\pi - x, & -\pi \le x < -\frac{\pi}{2} \end{cases} \]
Case 1: \( -\frac{\pi}{2} \le x \le \frac{\pi}{2} \)
\[ |\cos x| = x \]
Since RHS ≥ 0 ⇒ x ≥ 0
\[ \cos x = x \Rightarrow x \approx 0.739 \]
✔ One solution
Case 2: \( \frac{\pi}{2} < x \le \pi \)
\[ |\cos x| = \pi – x \Rightarrow -\cos x = \pi – x \Rightarrow \cos x = x – \pi \]
✔ One solution in this interval
Case 3: \( -\pi \le x < -\frac{\pi}{2} \)
\[ |\cos x| = -\pi – x \Rightarrow -\cos x = -\pi – x \Rightarrow \cos x = \pi + x \]
✔ One solution in this interval
Total solutions:
\[ 1 + 1 + 1 = 3 \]
Final Answer:
\[ \boxed{3} \]
Key Concept
Use identity \(1+\cos2x = 2\cos^2x\) and analyze piecewise behavior of \( \sin^{-1}(\sin x) \).