Question

sin2xsin4x=cos2x помогите

Answer 1

sin^4x-cos^4x=sin2xsin^4x-cos^4x=(sin^2x+cos^2x)*(sin^2x-cosx^2x) = 1*(sin^2x-cos^2x) = -cos2x-cos2x=sin2xsin2x+cos2x=0  делим на √21/√2sin2x+1/√2cos2x=0sin(2x+pi/4)=02x+pi/4=pi*k2x=pi*k-pi/4x=(pi*k)/2 - pi/8