If `4sin^4x+cos^4x=1,t h e nxi se q u a lto(n in Z)` `npi` (b) `npi+-sin^(-1)sqrt(2/5)` `(2npi)/3` (d) `2npi+-pi/4`

Options

A. `n pi`
B. `npi pm sin^(-1) sqrt(2/5)`
C. `(2 n pi)/3`
D. `2n pi pm pi/4`

Answer

A, B

Explanation

We have `4 sin^(4)x+cos^(4)x=1`
or `4 sin^(4)x=1-cos^(4)x`
`=(1-cos^(2) x) (1+ cos^(2) x)`
`=sin^(2) x (2-sin^(2) x)`
or `sin^(2)x[5 sin^(2) x-2]=0`
or `sin x=0` or `sin x= pm sqrt(2//5)`
`rArr x=n pi` or `x=n pi pm sin^(-1) sqrt(2/5), n in Z`