Problem
Solve: \( \sin^{-1}x = \frac{\pi}{6} + \cos^{-1}x \)
Solution
Step 1: Use identity
\[ \cos^{-1}x = \frac{\pi}{2} – \sin^{-1}x \]
Step 2: Substitute
\[ \sin^{-1}x = \frac{\pi}{6} + \left(\frac{\pi}{2} – \sin^{-1}x\right) \]
\[ \sin^{-1}x = \frac{2\pi}{3} – \sin^{-1}x \]
Step 3: Solve
\[ 2\sin^{-1}x = \frac{2\pi}{3} \]
\[ \sin^{-1}x = \frac{\pi}{3} \]
Step 4: Find x
\[ x = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \]
Final Answer
\[ \boxed{\frac{\sqrt{3}}{2}} \]
Explanation
Convert cos⁻¹x into sin⁻¹x and solve the equation algebraically.