Sin^3x-sin^2x cosx-3sinxcos^2x+3cos^3x=0

sin^2x (sinx - cosx) - 3cos^2x (sinx - cosx) = 0;

Вынесем (sinx - cosx) за скобки;

(sinx - cosx) (sin^2x - 3cos^2x) = 0;

1) sinx - cosx = 0;

Разделим на cosx;

cosx ≠ 0; = => x ≠ п/2 + пk, k∈Z;

tgx - 1 = 0;

tgx = 1, = => x = π/4 + πn, n∈z;

2) sin^2x - 3cos^2x = 0;

По формулам половинного угла получим:

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

1 - cos2x - 3 - 3cos2x = 0;

4cos2x = - 2;

cos2x = - 1/2; = => 2x = ± 2 п/3 + 2 пm, m∈Z.

x = ± п/3 + пm, m∈Z;

Ответ: x = π/4 + πn, n∈z; x = ± п/3 + пm, m∈Z;