Question

Solve the equation
${\sin }^{-1}x+{\sin }^{-1}2x=\frac{\pi }{3}$.

Answer 1

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

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

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

${\sin }^{-1}2x={\sin }^{-1}[\frac{\sqrt{3}}{2}\sqrt{1-x^2}-x\sqrt{1-{\left(\frac{\sqrt{3}}{2}\right)}^2}]$

${\sin }^{-1}2x={\sin }^{-1}[\frac{\sqrt{3}}{2}\sqrt{1-x^2}-x\sqrt{\frac{4-3}{2}}]$

${\sin }^{-1}2x={\sin }^{-1}[\frac{\sqrt{3}}{2}\sqrt{1-x^2}-\frac{x}{2}]$

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

$4x=\sqrt{3}\sqrt{1-x^2}-x$.

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

$(5x{)}^2=3(1-x^2)$

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

$28x^2=3$

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

$x=\pm \sqrt{\frac{3}{28}}=\pm \sqrt{\frac{3}{4\times 7}}$

$x=\pm \frac{\sqrt{3}}{2\sqrt{7}}$

$\therefore x=\frac{\sqrt{3}}{2\sqrt{7}},x=\frac{-\sqrt{3}}{2\sqrt{7}}$