Question

Find the value of x

If, ${\sin }^{-1}x+{\sin }^{-1}2x=\frac{\pi }{3}$

Answer 1

Given, ${\sin }^{-1}x+{\sin }^{-1}2x=\frac{\pi }{3}$

$\Rightarrow {\sin }^{-1}2x=\frac{\pi }{3}-{\sin }^{-1}x$

$\Rightarrow 2x=\sin (\frac{\pi }{3}-{\sin }^{-1}x)$

$\Rightarrow 2x=\sin \frac{\pi }{3}\cos \left({\sin }^{-1}x\right)-\cos \frac{\pi }{3}\sin \left({\sin }^{-1}x\right)$

$\Rightarrow 2x=\frac{\sqrt{3}}{2}\cos \left({\cos }^{-1}\sqrt{1-x^2}\right)-\frac{x}{2}$

$\Rightarrow 2x+\frac{x}{2}=\frac{\sqrt{3}}{2}\cos \left({\cos }^{-1}\sqrt{1-x^2}\right)$

$\Rightarrow \frac{4x+x}{2}=\frac{\sqrt{3}}{2}\cos \left({\cos }^{-1}\sqrt{1-x^2}\right)$

$\Rightarrow \frac{5x}{2}=\frac{\sqrt{3}}{2}\cos \left({\cos }^{-1}\sqrt{1-x^2}\right)$

$\Rightarrow 5x=\sqrt{3}\sqrt{1-x^2}$

squaring both sides, we get

$\Rightarrow 25x^2=3(1-x^2)$

$\Rightarrow 25x^2=3-3x^2$

$\Rightarrow 28x^2=3$

$\Rightarrow x^2=\frac{3}{28}$

$\therefore x=\pm \sqrt{\frac{3}{28}}$