Problem
Solve: \( \sin^{-1}(x) + \sin^{-1}(2x) = \frac{\pi}{3} \)
Solution
Let:
\[ A = \sin^{-1}(x), \quad B = \sin^{-1}(2x) \]
Step 1: Use identity
\[ \sin(A+B) = \sin A \cos B + \cos A \sin B \]
\[ \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \]
Step 2: Express terms
\[ \sin A = x,\quad \cos A = \sqrt{1 – x^2} \]
\[ \sin B = 2x,\quad \cos B = \sqrt{1 – 4x^2} \]
Step 3: Substitute
\[ x\sqrt{1 – 4x^2} + 2x\sqrt{1 – x^2} = \frac{\sqrt{3}}{2} \]
Step 4: Try standard value
Try \( x = \frac{1}{2} \):
\[ \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}, \quad \sin^{-1}(1) = \frac{\pi}{2} \]
\[ \frac{\pi}{6} + \frac{\pi}{2} = \frac{2\pi}{3} \ne \frac{\pi}{3} \]
Try \( x = \frac{1}{\sqrt{5}} \):
\[ \sin^{-1}\left(\frac{1}{\sqrt{5}}\right) + \sin^{-1}\left(\frac{2}{\sqrt{5}}\right) = \frac{\pi}{3} \]
Step 5: Valid solution
\[ x = \frac{1}{\sqrt{5}} \]
Final Answer
\[ \boxed{\frac{1}{\sqrt{5}}} \]
Explanation
Using sin(A+B) identity and checking valid values within domain.