Giải phương trình: $\tan x - 3\cot x = 4\left( {\sin x + \sqrt 3 \cos x} \right)$.

$\tan x - 3\cot x = 4\left( {\sin x + \sqrt 3 \cos x} \right)$

$\Leftrightarrow \frac{{\sin x}}{{\cos x}} - 3\frac{{\cos x}}{{\sin x}} = 4\left( {\sin x + \sqrt 3 \cos x} \right)\;\left( 1 \right)$

ĐK: $\left\{ \begin{array}{l}\cos x \ne 0\\\sin x \ne 0\end{array} \right. \Leftrightarrow x \ne \frac{{k\pi }}{2}\;\left( * \right)$

$\left( 1 \right) \Leftrightarrow \frac{{{{\sin }^2}x - 3{{\cos }^2}x}}{{\sin x\,.\,\cos x}} = 4\left( {\sin x + \sqrt 3 \cos x} \right)$

$\Leftrightarrow \left( {\sin x - \sqrt 3 \cos x} \right)\left( {\sin x + \sqrt 3 \cos x} \right) = 4\sin x\,.\,\cos x\,.\,\left( {\sin x + \sqrt 3 \cos x} \right)$

$\Leftrightarrow \left( {\sin x + \sqrt 3 \cos x} \right)\left( {\sin x - \sqrt 3 \cos x - 4\sin x\,.\,\cos x} \right) = 0$

$\Leftrightarrow \left( {\sin x + \sqrt 3 \cos x} \right)\left( {\sin x - \sqrt 3 \cos x - 2\sin 2x} \right) = 0$

$\Leftrightarrow \left[ \begin{array}{l}\sin x + \sqrt 3 \cos x = 0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( 2 \right)\\\sin x - \sqrt 3 \cos x - 2\sin 2x = 0\;\;\;\left( 3 \right)\end{array} \right.$

+) $\left( 2 \right) \Leftrightarrow \frac{1}{2}\sin x + \frac{{\sqrt 3 }}{2}\cos x = 0$

$\Leftrightarrow \sin x\,.\,\cos \frac{\pi }{3} + \cos x\,.\,\sin \frac{\pi }{3} = 0$

$\Leftrightarrow \sin \left( {x + \frac{\pi }{3}} \right) = 0$

$\Leftrightarrow x + \frac{\pi }{3} = k\pi$

$\Leftrightarrow x = - \frac{\pi }{3} + k\pi ,\;\left( {k \in \mathbb{Z}} \right)$

+) $\left( 2 \right) \Leftrightarrow \sin x - \sqrt 3 \cos x = 2\sin 2x$

$\Leftrightarrow \frac{1}{2}\sin x - \frac{{\sqrt 3 }}{2}\cos x = \sin 2x$

$\Leftrightarrow \sin x\,.\,\cos \frac{\pi }{3} - \cos x\,.\,\sin \frac{\pi }{3} = \sin 2x$

$\Leftrightarrow \sin \left( {x - \frac{\pi }{3}} \right) = \sin 2x$

$\Leftrightarrow \left[ \begin{array}{l}x - \frac{\pi }{3} = 2x + k2\pi \\x - \frac{\pi }{3} = \pi - 2x + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = - \frac{\pi }{3} + k2\pi \\x = \frac{{4\pi }}{9} + \frac{{k2\pi }}{3}\end{array} \right.\;\left( {k \in \mathbb{Z}} \right)$

Vậy phương trình trên có hai họ nghiệm là $S = \left\{ { - \frac{\pi }{3} + k\pi ;\;\frac{{4\pi }}{9} + \frac{{k2\pi }}{3},\;k \in \mathbb{Z}} \right\}$.