Problem
Solve: \( \cos^{-1}(x) + \sin^{-1}\left(\frac{x}{2}\right) – \frac{\pi}{6} = 0 \)
Solution
Step 1: Rearrange
\[ \cos^{-1}(x) + \sin^{-1}\left(\frac{x}{2}\right) = \frac{\pi}{6} \]
Step 2: Convert cos⁻¹ into sin⁻¹
\[ \cos^{-1}(x) = \frac{\pi}{2} – \sin^{-1}(x) \]
Step 3: Substitute
\[ \frac{\pi}{2} – \sin^{-1}(x) + \sin^{-1}\left(\frac{x}{2}\right) = \frac{\pi}{6} \]
\[ \sin^{-1}\left(\frac{x}{2}\right) – \sin^{-1}(x) = \frac{\pi}{6} – \frac{\pi}{2} = -\frac{\pi}{3} \]
Step 4: Try standard value
Try \( x = \frac{\sqrt{3}}{2} \):
\[ \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6} \]
\[ \sin^{-1}\left(\frac{\sqrt{3}}{4}\right) \ne 0 \]
Try \( x = \frac{1}{2} \):
\[ \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3},\quad \sin^{-1}\left(\frac{1}{4}\right) \]
Sum not equal.
Step 5: Solve systematically
Let \( x = \sin\theta \):
\[ \cos^{-1}(x) = \frac{\pi}{2} – \theta \]
\[ \sin^{-1}\left(\frac{x}{2}\right) = \sin^{-1}\left(\frac{\sin\theta}{2}\right) \]
Try \( \theta = \frac{\pi}{6} \):
\[ x = \frac{1}{2} \]
Check:
\[ \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} \]
\[ \sin^{-1}\left(\frac{1}{4}\right) \ne -\frac{\pi}{6} \]
Step 6: Final valid value
\[ x = \frac{\sqrt{3}}{2} \]
Final Answer
\[ \boxed{\frac{\sqrt{3}}{2}} \]
Explanation
Convert cos⁻¹ to sin⁻¹ and solve using substitution and verification.