If `x in [1, 0)`, then find the value of `cos^(-1) (2x^(2) - 1) - 2 sin^(-1) x`

Explanation

Let `x = cso theta, " for " x in [-1, 0), theta in (pi//2, pi]`
`rArr cos^(-1) (2x^(2) -1) - 2 sin^(-1) x`
`= cos^(-1) (2 cos^(2) theta -1) -2 sin^(-1) (cos theta)`
`= cos^(-1) (cos 2 theta) - 2 sin^(-1) (sin(pi//2 - theta))`
`= cos^(-1) (cos 2 theta) - 2 ((pi)/(2) - theta)`
`= 2 pi - 2 theta + 2 theta - pi`
`= pi`