Question

sin^2x-cos^4x=1 помогите

Answer 1

cos4x-sin2x=0cos4x = 1-2sin^2 2x1-2sin^2 (2x) -sin2x=02sin^2(2x)+sin2x-1=0D=9sin2x=1/2  -> x=((-1)^k   * pi) / 12 + (pi*l) /2sin2x=-1  => 2x= - pi/2+2pi*k  => x= - pi/4+pi*k