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

Options

A. `x in [1/sqrt2, 1]`
B. `x in [0, 1]`
C. `x in [1/sqrt3, 1]`
D. `x in [1/sqrt5, 1]`

Answer

A

Explanation

Let x = cos y, where `0 le y le pi, |x| le 1`
`2 cos^(-1) x = sin^(-1) (2 x sqrt(1 -x^(2)))`...(i)
`rArr 2 cos^(-1) (cos y) = sin^(-1) (2 cos y sqrt(1 - cos^(2) y))`
`= sin^(-1) (2 cos y sin y)`
`= sin^(-1) (sin 2 y)`
`rArr sin^(-1) (sin 2 y) = 2y " for " -pi//2 le y le pi//4`
and `2 cos^(-1) (cos y) = 2y " for " 0 le y le pi`
Thus, Eq. (i) holds only when
`y in [0, pi//4]`
`rArr x in [1//sqrt2, 1]`