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The equation `sin^4x+cos^4x+sin2x+alpha=0`is solvable for`-5/2lt=alphalt=1/2`(b) `-3lt=alpha

Answer» `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`
`2alpha=(y-1)^2-3`
`alpha=(y-1)^2/2-3/2`
`-1<=y<=1`
`-2<=y-1<=0`
`0<=(y-1)^2<=4`
`-3<=(y-1)^2-3<=1`
`-3<=2alpha<=1`
`-3/2<=alpha<=1/2`.


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