sin3x-sinx/1+cosx=0 решить уравнение

1 - cosx ≠ 0;

cosx ≠ 1;

x ≠ 2πk, k ∈ Z.

(sinx - sin3x)/(1 - cosx) = 0;

sinx - sin3x = 0;

sin3x - sinx = 0;

2sin((3x - x)/2) * cos((3x + x)/2) = 0;

2sinx * cos2x = 0;

[sinx = 0;

[cos2x = 0;

[x = πk, k ∈ Z;

[2x = π/2 + πk, k ∈ Z;

[x = πk, k ∈ Z;

[x = π/4 + πk/2, k ∈ Z.

{x ≠ 2πk, k ∈ Z;

{[x = πk, k ∈ Z;

{[x = π/4 + πk/2, k ∈ Z;

[x = π + 2πk, k ∈ Z;

[x = π/4 + πk/2, k ∈ Z.

Ответ: π + 2πk; π/4 + πk/2, k ∈ Z.