2. Решить уравнение: 1) tg2x = √3/3; 2) sin5x = sin3x; 3) (cosx + √3/2) (tgx - √3) = 0

Решение:

1. \( \operatorname{tg} 2x = \frac{\sqrt{3}}{3} \)
   \( 2x = \frac{\pi}{6} + \pi n, n \in \mathbb{Z} \)
   \( x = \frac{\pi}{12} + \frac{\pi n}{2}, n \in \mathbb{Z} \)
2. \( \sin 5x = \sin 3x \)
   \( 5x = 3x + 2\pi k, k \in \mathbb{Z} \) или \( 5x = \pi - 3x + 2\pi k, k \in \mathbb{Z} \)
   \( 2x = 2\pi k \implies x = \pi k, k \in \mathbb{Z} \)
   \( 8x = \pi + 2\pi k \implies x = \frac{\pi}{8} + \frac{\pi k}{4}, k \in \mathbb{Z} \)
3. \( \left( \cos x + \frac{\sqrt{3}}{2} \right) (\operatorname{tg} x - \sqrt{3}) = 0 \)
   \( \cos x + \frac{\sqrt{3}}{2} = 0 \) или \( \operatorname{tg} x - \sqrt{3} = 0 \)
   \( \cos x = -\frac{\sqrt{3}}{2} \) \(\implies x = \pm \frac{5\pi}{6} + 2\pi n, n \in \mathbb{Z}\)
   \( \operatorname{tg} x = \sqrt{3} \) \(\implies x = \frac{\pi}{3} + \pi m, m \in \mathbb{Z}\)

Ответ: 1) \( x = \frac{\pi}{12} + \frac{\pi n}{2}, n \in \mathbb{Z} \); 2) \( x = \pi k, k \in \mathbb{Z} \) или \( x = \frac{\pi}{8} + \frac{\pi k}{4}, k \in \mathbb{Z} \); 3) \( x = \pm \frac{5\pi}{6} + 2\pi n, n \in \mathbb{Z} \) или \( x = \frac{\pi}{3} + \pi m, m \in \mathbb{Z} \).