The equation `sin^4x+cos^4x+sin2x+alpha=0` is solvable for

Options

A. `-5/2lt=alphalt=1/2`
B. `-3lt=alpha<1`
C. `-3/2lt=alphalt=1/2`
D. `-1lt=alphalt=1`

Explanation

`sin^4x+cos^4x+sin2x+alpha=0`

`(sin^2x)^2+(cos^2x)^2+sin2x+alpha=0`

`(sinn^2x+cos^2x)^2-2sin^2x*cos^2x+sin2x+alpha=0`

`1-2sin^2x*cos^2x+sin2x+alpha=0`

`1-2(sin^2 2x)/4+sin2x+alpha=0`

`2-sin^2 2alpha+2sin2x+2alpha=0`

`2alpha=sin^2 2alpha-2sin2x-2`

`2alpha=y^2-2y-2`

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