Question

!!! ГЛАВНЫЙ МОЗГ,ПОМОГИ,35 БАЛЛОВ !!!​

Answer 1

Ответ:

$3$

Объяснение:

$\sqrt{3}tg2x+3=0;$

$\sqrt{3}tg2x=-3;$

$tg2x=-\frac{3}{\sqrt{3}};$

$tg2x=-\sqrt{3};$

$2x=arctg(-\sqrt{3})+\pi n, \quad n \in \mathbb {Z};$

$2x=-arctg(\sqrt{3})+\pi n, \quad n \in \mathbb {Z};$

$2x=-\frac{\pi}{3}+\pi n, \quad n \in \mathbb {Z};$

$x=-\frac{\pi}{6}+\frac{\pi}{2} n, \quad n \in \mathbb {Z};$

$[\frac{\pi}{3}; \frac{3\pi}{2}]=[\frac{2\pi}{6}; \frac{9\pi}{6}];$

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

$\frac{2\pi}{6} \leq -\frac{\pi}{6}+\frac{3\pi}{6}n \leq \frac{9\pi}{6} \quad | \quad +\frac{\pi}{6}$

$\frac{3\pi}{6} \leq \frac{3\pi}{6}n \leq \frac{10\pi}{6} \quad | \quad \cdot \frac{6}{3\pi}$

$1 \leq n \leq \frac{10}{3};$

$1 \leq n \leq \frac{10}{3}, \quad n \in \mathbb {Z} \Rightarrow n \in [1; 3];$

Количество корней: 3.