Решим уравнение $$cos 3x = \frac{\sqrt{3}}{2}$$.
$$3x = \pm arccos(\frac{\sqrt{3}}{2}) + 2\pi n, n \in Z$$
$$3x = \pm \frac{\pi}{6} + 2\pi n, n \in Z$$
$$x = \pm \frac{\pi}{18} + \frac{2\pi n}{3}, n \in Z$$
Ответ: $$x = \pm \frac{\pi}{18} + \frac{2\pi n}{3}, n \in Z$$