If `4sin^4x+cos^4x=1`,then `x` is equal to `(n in Z)` (a)`npi` (b) `npi+-sin^(-1)sqrt(2/5)` (c)`(2npi)/3` (d) `2npi+-pi/4`

Explanation

`4sin^4x+cos^4x = 1`

`=>4sin^4x -(1-cos^4x) = 0`

`=>4sin^4x - (1-cos^2x)(1+cos^2x) = 0`

`=>4sin^4x - sin^2x(1+cos^2x) = 0`

`=>sin^2x(4sin^2x - (1+(1-sin^2x)) = 0`

`=>sin^2x(5sin^2x - 2) = 0`

`=>sin^2x = 0 or 5sin^2x - 2 =0`

`=>sinx = 0 or sin x = +-sqrt(2/5)`

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