Solve `sin^(-1)x-cos^(-1)x=sin^(-1)(3x-2)`

Explanation

`sin^(-1) x - cos^(-1) x = sin^(-1) (3x -2)`
This equation is valid if `-1 le x le 1 and -1 le 3x - 2 le 1`
`rArr x in [(1)/(3), 1]`
given equation can be rewritten as:
`(pi)/(2) - cos^(-1) x = cos^(-1) x = (pi)/(2) - cos^(-1) (3x -2)`
`rArr 2 cos^(-1) x = cos^(-1) (3 x -2)`
`rArr cos^(-1) (2x^(2) - 1) = cos^(-1) (3x -2)`
`rArr 2x^(2) -1 = 3x - 2`
`rArr 2x^(2) - 3x + 1 = 0`
`rArr x = 1 " or " (1)/(2)`