Sin3x*cos3x=-1/2 sinx-sin3x+sin5x=0 решите уровнение

1. Синус двойного угла:

sin2a = 2sina * cosa; sin3x * cos3x = - 1/2; 2sin3x * cos3x = - 1; sin6x = - 1; 6x = - π/2 + 2πk, k ∈ Z; x = - π/12 + πk/3, k ∈ Z.

2. Сумма синусов:

sina + sinb = 2sin ((a + b) / 2) * cos ((a - b) / 2); sinx - sin3x + sin5x = 0; (sin5x + sinx) - sin3x = 0; 2sin ((5x + x) / 2) * cos ((5x - x) / 2) - sin3x = 0; 2sin3x * cos2x - sin3x = 0; 2sin3x (cos2x - 1/2) = 0; [sin3x = 0;

[cos2x - 1/2 = 0; [sin3x = 0;

[cos2x = 1/2; [3x = πk, k ∈ Z;

[2x = ±π/3 + 2πk, k ∈ Z; [x = πk/3, k ∈ Z;

[x = ±π/6 + πk, k ∈ Z.

Ответ:

1) - π/12 + πk/3, k ∈ Z; 2) πk/3; ±π/6 + πk, k ∈ Z.