Question

Solve for $x : sin^{-1}2x + sin^{-1}3x = \frac{\pi}{3}$

• A. $ \sqrt{\frac{76}{3}}$
• B. $ \sqrt{\frac{3}{76}}$
• C. $ \frac{3}{\sqrt{76}}$
• D. $\frac{\sqrt{3}}{76}$

Answer 1

Answer: (B)

$sin^{-1}\,2x + sin^{-1}\,3x = \frac{\pi}{3}$
$\Rightarrow \frac{\pi }{2} - cos^{-1} 2x + \frac{\pi }{2} - cos^{-1} 3x = \frac{\pi }{3}$
$\Rightarrow cos^{-1} 2x + cos^{-1} 3x = \frac{2\pi }{3}$
$\Rightarrow cos^{-1} \left\{6x^{2} - \sqrt{1-\left(2x\right)^{2} } \sqrt{1-\left(3x\right)^{2}}\right\} = \frac{2\pi }{3}$
$\Rightarrow 6x^{2} - \sqrt{\left(1-13x^{2}+36x^{4}\right)} = -\frac{1}{2}$
$\Rightarrow \left(6x^{2}+\frac{1}{2}\right)^{2} = 1 - 13x^{2} + 36x^{4}$
$\Rightarrow 19x^{2} = \frac{3}{4}$
$\Rightarrow x = \pm \sqrt{\frac{3}{76}}$
But sum of two negative numbers cannot be $\frac{\pi}{3}$.
$\therefore x = \sqrt{\frac{3}{76}}$ is the only solution.