3tanx+√3cotx -3 -√3 =0

Đáp án:

$x=\left\{{\dfrac{\pi}4+k\pi;\dfrac{\pi}6+k\pi}\right\}$ $(k\in\mathbb Z)$

Lời giải:

ĐKXĐ: $\sin x\neq0, \cos x \neq 0$

$\Leftrightarrow\sin2x\ne0\Leftrightarrow x\ne\dfrac{k\pi}2$ $(k\in\mathbb Z)$

$3\tan x+\sqrt3\cot x-3-\sqrt3=0$

$⇔ 3(\tan x-1)+ \sqrt3(\cot x-1)=0$

$⇔ 3\left({ \dfrac{\sin x }{\cos x }-1}\right) + \sqrt3\left({\dfrac{\cos x }{\sin x }-1}\right)=0$

$⇔ 3.\dfrac{\sin x -\cos x }{\cos x }+\sqrt3 \dfrac{\cos x -\sin x }{\sin x }=0$

$⇔ (\sin x -\cos x )\left({ \dfrac{3}{\cos x } -\dfrac{\sqrt3}{\sin x } }\right)=0$

$⇔ (\sin x -\cos x )(3\sin x - \sqrt3\cos x ). \dfrac{1}{\sin x \cos x }=0$

$⇔ \sin x -\cos x =0$ (1) hoặc $3\sin x - \sqrt3\cos x =0$ (2)

(1) $\Leftrightarrow\sin\left({x-\dfrac{\pi}4}\right)=0$

$\Leftrightarrow x-\dfrac{\pi}4=k\pi\Leftrightarrow x=\dfrac{\pi}4+k\pi$ $(k\in\mathbb Z)$

(2) $\Leftrightarrow\sqrt3\left({\dfrac{\sqrt3}2\sin x-\dfrac12\cos x}\right)=0$

$\Leftrightarrow \sin\left({x-\dfrac{\pi}6}\right)=0$

$\Leftrightarrow x-\dfrac{\pi}6=k\pi\Leftrightarrow x=\dfrac{\pi}6+k\pi$ $(k\in\mathbb Z)$

Vậy phương trình có nghiệm 

$x=\left\{{\dfrac{\pi}4+k\pi;\dfrac{\pi}6+k\pi}\right\}$ $(k\in\mathbb Z)$.