Solve `sin^(-1)(1-x)-2s in^(-1)x=pi/2`

Explanation

Given , `sin^(-1) (1 - x) - 2 sin^(-1) x = pi//2`
Let `x sin y`
Therefore, the given equation reduces sto
`sin^(-1) (1 - sin y) - 2 sin^(-1) sin y= (pi)/(2)`
or `sin^(-1) (1 - sin y) - 2 y = (pi)/(2)`
or `sin^(-1) (1 - sin y) = (pi)/(2) + 2y`
or `1 - sin y = sin ((pi)/(2) + 2y)`
or `1 - sin y = cos 2 y`
or `1 - cos 2y = sin y`
or `2 sin^(2) y = sin y`
or `2 sin^(2) y - sin y = 0`
or `sin y (2 sin y - 1) = 0`
`rArr sin y = 0 " or " 1//2`
`rArr x = 0 " or " 1//2`
But, when `x = (1)/(2)`, it can be observed that
`L.H.S. = sin^(-1). (1 - (1)/(2)) - 2 sin ^(-1). (1)/(2)`
`= sin^(-1). ((1)/(2)) - 2 sin^(-1). (1)/(2)`
`= - sin^(-1). (1)/(2)`
`= - (pi)/(6) != (pi)/(2)`
Therefore, `x = (1)/(2)` is not the solution of the given equation Thus, `x = 0`