решите уравнение sin^3x-sin^2xcosx-3sinxcos^2x+3cos^3x=0

sin²x(sinx-cosx)-3cos²x(sinx-cosx)=0(sinx-cosx)(sin²x-3cos²x=0sinx-cosx=0/cosxtgx-1=0tgx=1⇒x=π/4+πn,n∈zsin²x-3cos²x=0(1-cos2x)/2-3(1+cos2x)/2=01-cos2x-3-3cos2x=0-4cos2x=2cos2x=-1/22x=+-2π/3+2πkx=+-π/3+πk,k∈z