Решим уравнение $$sin(2x - \frac{\pi}{3}) + 1 = 0$$.
$$sin(2x - \frac{\pi}{3}) = -1$$.
$$2x - \frac{\pi}{3} = arcsin(-1) + 2\pi n, n \in \mathbb{Z}$$.
$$2x - \frac{\pi}{3} = -\frac{\pi}{2} + 2\pi n, n \in \mathbb{Z}$$.
$$2x = -\frac{\pi}{2} + \frac{\pi}{3} + 2\pi n, n \in \mathbb{Z}$$.
$$2x = -\frac{\pi}{6} + 2\pi n, n \in \mathbb{Z}$$.
$$x = -\frac{\pi}{12} + \pi n, n \in \mathbb{Z}$$.
Ответ: $$x = -\frac{\pi}{12} + \pi n, n \in \mathbb{Z}$$.