Решите уравнение 2/(1-cos(2x+pi/3))=sin(x+pi/6)

Вопрос пользователя:

$2 = sin(x + frac{pi}{6})(1 - cos({2x + frac{pi}{3}))$

$2 = sin(x+frac{pi}{6}) - sin(x + frac{pi}{6})cos2(x + frac{pi}{6})$

$2 = sin(x + frac{pi}{6}) - sin(x+frac{pi}{6})(1 - sin^{2}(x+frac{pi}{6}))$

$2 = sin(x + frac{pi}{6}) - sin(x + frac{pi}{6}) - 2sin^{3}(x + frac{pi}{6})$

$2sin^{3}(x + frac{pi}{6}) = 2$

$sin(x + frac{pi}{6}) = 1$

$x + frac{pi}{6} = frac{pi}{2} + 2pi k$

$x = frac{pi}{2} - frac{pi}{6} + 2pi k$

$x = frac{pi}{3} + 2pi k$ , k ∈ Z