Question

Sin(П/2+3х)cos2х-1=sin3xcos(3П/2-2x)

Answer 1

$sin(frac{pi}{2}+3x)cos2x-1=sin3xcos(frac{3pi }{2}-2x)\\cos3xcdot cos2x-(-sin3xcdot sin2x)=1\\cos3xcdot cos2x+sin3xcdot sin2x=1\\cos alpha cdot cos beta +sin alpha cdot sin beta =cos( alpha - beta )\\cos(3x-2x)=1\\cosx=1\\x=2pi n,; nin Z$