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. If sin x + sin2x + sin3x= 1, then cos6x-4cos4x+8cos2x = ?
sinx+ (1-cos2x)+sinx(1-cos2x)=1
sinx(1+1-cos2x)+1-cos2x=1
sinx(1+1-cos2x)-cos2x=1-1=0
sinx(2-cos2x)-cos2x=0
2-cos2x = cos2x / sinx [sinx =√1-cos2x]
2-cos2x = cos2x /√1-cos2x
squaring on both sides
(2-cos2x)2 = (cos2x)2 /(√1-cos2x)2
(4+cos4x-4cos2x)(1-cos2x) = cos4x
4+cos4x-4cos2x-4cos2x-cos6x+4cos4x= cos4x
4-8cos2x-cos6x+4cos4x=0
on rearranging
8cos2x+cos6x-4cos4x=4
hence solved
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