Đkxđ:
$\begin{cases} \cos x \ne 0 ⇔ x \ne \dfrac{\pi}{2}+mπ \\ \sin x \ne 0 ⇔ x \ne mπ \end{cases}$
Ta có:
`\sqrt{3}tanx-6cotx+2\sqrt{3}-3=0`
⇔ `\sqrt{3}tanx-\frac{6}{tanx}+2\sqrt{3}-3=0`
⇔ `\sqrt{3}tan^2x-6+(2\sqrt{3}-3)tanx=0`
⇔ $\left [\begin{array}{l} \tan x=\sqrt{3} \\ \tan x=-2 \end{array} \right.$
⇔ $\left [\begin{array}{l} x=\dfrac{\pi}{3}+kπ \\ x=\arctan(-2)+kπ \end{array} \right.$ (thỏa)