Question

${\sin }^{-1}(1-x)-2{\sin }^{-1}x=\frac{\pi }{2}$, then $x$ is equal to:

Answer 1

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

$⟹{\sin }^{-1}(1-x)=\frac{\pi }{2}+2{\sin }^{-1}x$

$⟹(1-x)=\sin (\frac{\pi }{2}+2{\sin }^{-1}x)$

$⟹(1-x)=\cos \left(2{\sin }^{-1}x\right)$

$⟹(1-x)=\cos \left({\cos }^{-1}\right(1-2x^2\left)\right)$

$⟹1-x=1-2x^2$

$⟹2x-x=0$

$⟹x(2x-1)=0$

So, $x=0$ or $x=\frac{1}{2}$

if $x=\frac{1}{2}$

${\sin }^{-1}\left(\frac{1}{2}\right)-2{\sin }^{-1}\left(\frac{1}{2}\right)=-\frac{\pi }{6}$ {does not satisfy the equation}

$\therefore x=0$