0353. a) 2cos()=√3; 6) √3tg + = 3;

353. a) Решим уравнение $$2\cos\left(\frac{x}{2} + \frac{\pi}{6}\right) = \sqrt{3}$$.

$$\cos\left(\frac{x}{2} + \frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$$

$$\frac{x}{2} + \frac{\pi}{6} = \pm \arccos \frac{\sqrt{3}}{2} + 2\pi k, k \in \mathbb{Z}$$

$$\frac{x}{2} + \frac{\pi}{6} = \pm \frac{\pi}{6} + 2\pi k, k \in \mathbb{Z}$$

$$\frac{x}{2} = -\frac{\pi}{6} \pm \frac{\pi}{6} + 2\pi k, k \in \mathbb{Z}$$

$$x = -\frac{\pi}{3} \pm \frac{\pi}{3} + 4\pi k, k \in \mathbb{Z}$$

$$x_1 = -\frac{\pi}{3} + \frac{\pi}{3} + 4\pi k = 4\pi k, k \in \mathbb{Z}$$

$$x_2 = -\frac{\pi}{3} - \frac{\pi}{3} + 4\pi k = -\frac{2\pi}{3} + 4\pi k, k \in \mathbb{Z}$$

Ответ: $$x = 4\pi k, x = -\frac{2\pi}{3} + 4\pi k, k \in \mathbb{Z}$$

353. б) Решим уравнение $$\sqrt{3}\operatorname{tg}\left(\frac{x}{3} + \frac{\pi}{6}\right) = 3$$.

$$\operatorname{tg}\left(\frac{x}{3} + \frac{\pi}{6}\right) = \frac{3}{\sqrt{3}} = \sqrt{3}$$

$$\frac{x}{3} + \frac{\pi}{6} = \operatorname{arctg} \sqrt{3} + \pi k, k \in \mathbb{Z}$$

$$\frac{x}{3} + \frac{\pi}{6} = \frac{\pi}{3} + \pi k, k \in \mathbb{Z}$$

$$\frac{x}{3} = \frac{\pi}{3} - \frac{\pi}{6} + \pi k, k \in \mathbb{Z}$$

$$\frac{x}{3} = \frac{\pi}{6} + \pi k, k \in \mathbb{Z}$$

$$x = \frac{\pi}{2} + 3\pi k, k \in \mathbb{Z}$$

Ответ: $$x = \frac{\pi}{2} + 3\pi k, k \in \mathbb{Z}$$