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

Sin²x (sinx-cosx) - 3cos²x (sinx-cosx) = 0

(sinx-cosx) (sin²x-3cos²x=0

sinx-cosx=0/cosx

tgx-1=0

tgx=1⇒x=π/4+πn, n∈z

sin²x-3cos²x=0

(1-cos2x) / 2-3 (1+cos2x) / 2=0

1-cos2x-3-3cos2x=0

-4cos2x=2

cos2x=-1/2

2x=+-2π/3+2πk

x=+-π/3+πk, k∈z