Решите уравнения: A) $$sinx = -\frac{1}{2}$$ Б) $$cos(\frac{\pi}{6} - \frac{2x}{3}) = -\frac{\sqrt{3}}{2}$$

A) $$sinx = -\frac{1}{2}$$ $$x = arcsin(-\frac{1}{2}) + 2\pi n$$, $$x = -\frac{\pi}{6} + 2\pi n$$ $$x = \pi - arcsin(-\frac{1}{2}) + 2\pi n$$, $$x = \pi + \frac{\pi}{6} + 2\pi n = \frac{7\pi}{6} + 2\pi n$$ Б) $$cos(\frac{\pi}{6} - \frac{2x}{3}) = -\frac{\sqrt{3}}{2}$$ $$\frac{\pi}{6} - \frac{2x}{3} = arccos(-\frac{\sqrt{3}}{2}) + 2\pi n$$, $$\frac{\pi}{6} - \frac{2x}{3} = \frac{5\pi}{6} + 2\pi n$$ $$-\frac{2x}{3} = \frac{5\pi}{6} - \frac{\pi}{6} + 2\pi n = \frac{4\pi}{6} + 2\pi n = \frac{2\pi}{3} + 2\pi n$$ $$x = -\pi - 3\pi n$$ $$\frac{\pi}{6} - \frac{2x}{3} = -arccos(-\frac{\sqrt{3}}{2}) + 2\pi n$$, $$\frac{\pi}{6} - \frac{2x}{3} = -\frac{5\pi}{6} + 2\pi n$$ $$-\frac{2x}{3} = -\frac{5\pi}{6} - \frac{\pi}{6} + 2\pi n = -\frac{6\pi}{6} + 2\pi n = -\pi + 2\pi n$$ $$x = \frac{3\pi}{2} - 3\pi n$$