Problem
Solve: \( 4\sin^{-1}x = \pi – \cos^{-1}x \)
Solution
Step 1: Use identity
\[ \cos^{-1}x = \frac{\pi}{2} – \sin^{-1}x \]
Step 2: Substitute
\[ 4\sin^{-1}x = \pi – \left(\frac{\pi}{2} – \sin^{-1}x\right) \]
\[ 4\sin^{-1}x = \frac{\pi}{2} + \sin^{-1}x \]
Step 3: Solve
\[ 3\sin^{-1}x = \frac{\pi}{2} \]
\[ \sin^{-1}x = \frac{\pi}{6} \]
Step 4: Find x
\[ x = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \]
Step 5: Check validity
Since \( \sin^{-1}x \in [-\frac{\pi}{2}, \frac{\pi}{2}] \), the solution is valid.
Final Answer
\[ \boxed{\frac{1}{2}} \]
Explanation
Convert cos⁻¹x into sin⁻¹x, reduce to a linear equation, and solve.