Question

$si{n}^{-1}(1-x)-2si{n}^{-1}x=\frac{\pi }{2}$ then find the value of $x$ .

Answer 1

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

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

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

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

$1-x=cos\left[co{s}^{-1}\right(1-2x^2\left)\right]$

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

$2x^2-x=0$

$x(2x-1)=0$

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

If $x=\frac{1}{2}$, then the equation is not satisfied.

Hence, $x=0$.