Решить sin2x/1-cosx=2sinx

Sin2x / (1 - cosx) = 2sinx

[sin2x - 2sinx + 2sinxcosx] / (1 - cosx) = 0

[2sinxcosx - 2sinx + 2sinxcosx] / (1 - cosx) = 0

[4sinxcosx - 2sinx] / (1 - cosx) = 0

sinx * (4cosx - 2) = 0,

ОДЗ: 1 - cosx ≠ 0, cosx ≠ 1, x ≠ 2πk, k∈Z

1) sinx = 0

x = πn, n∈Z

2) 4cosx - 2 = 0

cosx = 1/2

x = (+ -) arccos (1/2) + 2πk, n∈Z

x = (+ -) * (π/3) + 2πk, k∈Z