If `2sin^2((pi/2)cos^2x)=1-cos(pis in2x),x!=(2n+1)pi/2,n in I ,` then `cos2x` is equal to `1/5` (b) `3/5` (c) `4/5` (d) `1`

Options

A. `x=(2n+1) pi/2, n in Z`
B. `tan x=1/2, n in Z`
C. `tan x=-1/2, n in Z`
D. `x=(n pi)/2, n in Z`

Answer

A, B, C

Explanation

`2 sin^(2)(pi/2 cos^(2) x)=1-cos (pi sin 2x)`
`rArr 2 sin^(2) {(pi/2)cos^(2) x}=2 sin^(2) {(pi/2) sin 2x}`
`rArr (pi/2) cos^(2) x= pm (pi/2) sin 2x`
`rArr cos x=0` or `cos x = pm 2 sin x`
`rArr x=(2n+1) pi/2, n in Z` or `tan x = pm 1/2`