Question

The number of solutions of $\sin^{- 1}x + \sin^{- 1}2x = \frac{\pi}{3}$ is

• A. 0
• B. 1
• C. 2
• D. Infinitie

Answer 1

Answer: (B)

1

Answer 2

$\sin^{- 1}2x = \sin^{- 1}\frac{\sqrt{3}}{2} - \sin^{- 1}x = \sin^{- 1}\left\lbrack \frac{\sqrt{3}}{2}.\sqrt{1 - x^{2}} - x\sqrt{1 - \frac{3}{4}} \right\rbrack$

$\therefore$ $2x = \frac{\sqrt{3}}{2}\sqrt{1 - x^{2}} - \frac{x}{2}$

$\therefore\left( \frac{5x}{2} \right)^{2} = \frac{3}{4}(1 - x^{2})$ or $28x^{2} = 3$ $\Rightarrow$ $x = \sqrt{\frac{3}{28}} = \frac{1}{2}\sqrt{\frac{3}{7}}$,

(not $- \frac{1}{2}\sqrt{\frac{3}{7}}$)