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.
