`"cos" 2x =(sqrt(2) + 1) ("cos"x- (1)/(sqrt(2))),"cos" x ne (1)/(2) rArr x in `

Options

A. `{2n pi +-(pi)/(3): n in Z}`
B. `{2npi +-(pi)/(6): n inZ}`
C. `{2npi + - (pi)/(2): n inZ}`
D. `{2npi +- (pi)/(4): n inZ}`

Answer

D

Explanation

We have,
`"cos" 2x = (sqrt(2) + 1) ("cos" x - (1)/(sqrt(2)))`
`(2 "cos"^(2) x-1) - ((sqrt(2) + 1))/(sqrt(2)) (sqrt(2)"cos"x-1) =0`
`rArr (sqrt(2) "cos" x - 1){(sqrt(2) "cos" x + 1) - ((sqrt(2) + 1)/(sqrt(2)))} = 0`
`rArr sqrt(2) "cos" x - 1 = 0 "or", sqrt(2)"cos" x - (1)/(sqrt(2)) = 0`
`rArr "cos"x = (1)/(sqrt(2)) "or, cos" x = (1)/(2)`
But, `"cos" x ne (1)/(2)`
`therefore "cos"x = (1)/(sqrt(2)) rArr "cos" x = "cos" (pi)/(4) rArr x = 2n pi +-(pi)/(4), n in Z`