(sin2x+sin пи/6)(cos2x-1)=0

Question

(sin2x+sin пи/6)(cos2x-1)=0

Answer 1

(sin2x+sin п/6) (cos2x-1)= (sin2x+sin п/6) *cos2x- (sin2x+cos2x) =
=(sin2x+sin п/6)*cos2x-sin2x-cos2x = (sin2x+sin п/6)-sin2x =
= sin2x+sin п/6-sin2x = sin п/6 = sin 30 = 1/2