Sin2xsin4x=cos2x помогите

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