Đáp án:
$S=\left\{\dfrac{\pi}{6}+k\pi;-\dfrac{\pi}{3}+k\pi\,\bigg{|}\,k\in\mathbb Z\right\}$
Giải thích các bước giải:
ĐKXĐ: $x\ne k\pi$
$\sqrt 3\cot^2x-2\cot x-\sqrt 3=0$
$⇔\sqrt 3\cot^2x-3\cot x+\cot x-\sqrt 3=0$
$⇔\sqrt 3\cot x(\cot x-\sqrt 3)+(\cot x-\sqrt 3)=0$
$⇔(\cot x-\sqrt 3)(\sqrt 3\cot x+1)=0$
$⇔\left[ \begin{array}{l}\cot x-\sqrt 3=0\\\sqrt 3\cot x+1=0\end{array} \right.⇔\left[ \begin{array}{l}\cot x=\sqrt 3\\\cot x=-\dfrac{1}{\sqrt 3}\end{array} \right.$
$⇔\left[ \begin{array}{l}\tan x=\dfrac{\sqrt 3}{3}\\\tan x=-\sqrt 3\end{array} \right.⇔\left[ \begin{array}{l}x=\dfrac{\pi}{6}+k\pi\\x=-\dfrac{\pi}{3}+k\pi\end{array} \right.\,\,(k\in\mathbb Z)$
Vậy $S=\left\{\dfrac{\pi}{6}+k\pi;-\dfrac{\pi}{3}+k\pi\,\bigg{|}\,k\in\mathbb Z\right\}$.
