Решить уравнения ... Sinx+sin2x-sin3x=o

По формулам тригонометрических тождеств:

sin x + sin 2x = 2 sin (3/2 x) cos (1/2 x),

sin 3x = 2 sin (3/2 x) cos (3/2 x),

cos (3/2 x) = 4 cos³ (1/2 x) - 3 cos (1/2 x).

Преобразуем уравнение:

2 sin (3/2 x) cos (1/2 x) - 2 sin (3/2 x) cos (3/2 x) = 0,

2 sin (3/2 x) ⋅ (cos (1/2 x) - (4 cos³ (1/2 x) - 3 cos (1/2 x))) = 0,

2 sin (3/2 x) ⋅ 4 cos (1/2 x) ⋅ (1 - cos² (1/2 x)) = 0.

sin (3/2 x) = 0,

x = (2/3) πn, n ∊ ℤ.

cos (1/2 x) = 0,

x = π (2n + 1), n ∊ ℤ.

1 - cos² (1/2 x) = 0,

cos (1/2 x) = ±1,

x = 2πn, n ∊ ℤ.

Ответ: x₁ = (2/3) πn, n ∊ ℤ; x₂ = πn, n ∊ ℤ.