$\sqrt 3 \cot \alpha - 3 = 0 $
$⇔ \sqrt 3 \cot \alpha = 3$
$⇔ \cot \alpha = \sqrt 3$
$⇔ \alpha = \dfrac{\pi}{6}+k\pi \ (k \in \mathbb{Z})$
Ta có: $0 \le \dfrac{\pi}{6}+k\pi \le 2018\pi$
$⇔ 0 - \dfrac{\pi}{6} \le k\pi \le 2018\pi - \dfrac{\pi}{6}$
$⇔ - \dfrac{\pi}{6} \le k\pi \le \dfrac{12107\pi}{6}$
$⇔ -\dfrac{1}{6} \le k \le 2017,83...$
$⇒ k \in \{0;1;2;3;4;5;6;...;2017\}$
$⇒ k$ có $2018$ nghiệm
Vậy $\sqrt 3 \cot \alpha - 3 = 0$ có $2018$ nghiệm
