Question

Find the value of x which satisfy equation : ${\sin }^{-1}x+{\sin }^{-1}2x=\frac{\pi }{3}$.

• A. $x=\frac{1}{2}\sqrt{\frac{3}{7}}$
• B. $x=\frac{1}{3}\sqrt{\frac{4}{7}}$
• C. $x=\frac{1}{3}\sqrt{\frac{3}{7}}$
• D. $x=\frac{1}{2}\sqrt{\frac{4}{7}}$

Answer 1

The correct option is A $x=\frac{1}{2}\sqrt{\frac{3}{7}}$

Given, ${\sin }^{-1}x+{\sin }^{-1}2x=\frac{\pi }{3}$ (i)
${\sin }^{-1}2x={\sin }^{-1}\frac{\sqrt{3}}{2}-{\sin }^{-1}x={\sin }^{-1}[\frac{\sqrt{3}}{2}\sqrt{1-x^2}-x\sqrt{1-\frac{3}{4}}]$
$2x=\frac{\sqrt{3}}{2}\sqrt{1-x^2}-\frac{x}{2}$
${\left(\frac{5x}{2}\right)}^2=\frac{3}{4}(1-x^2)$

$28x^2=3$
$x=\sqrt{\frac{3}{28}}=\frac{1}{2}\sqrt{\frac{3}{7}}$
($\because x=-\frac{1}{2}\sqrt{\frac{3}{7}}$ makes
L.H.S. of Eq. (i) negative)