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

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

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

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

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

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

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

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

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

$x = \frac{\pi}{3} + 2\pi k$ , k ∈ Z