Question

Solve: $si{n}^{-1}(1-x)-2{\sin }^{-1}x=\frac{\pi }{2}$.

Answer 1

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

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

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

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

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

$1-x=1-2x^2$

$x=2x^2$

$x(x-\frac{1}{2})=0$

Hence, $x=0,x=\frac{1}{2}$

But when $x=\frac{1}{2}$ in the equation, we get

$\frac{\pi }{6}\ne \frac{\pi }{2}$

Hence, $x=0$