Điều kiện: $\left\{ \begin{array}{l}\cos x \ne 0\\\cos \left( {x + \dfrac{\pi }{3}} \right) \ne 0\\\cos \left( {x + \dfrac{{2\pi }}{3}} \right) \ne 0\end{array} \right.$
${\rm{pt}}$\( \Leftrightarrow \tan x + \tan \left( {x + \dfrac{\pi }{3}} \right) + \tan \left( {x + \dfrac{{2\pi }}{3}} \right)\)\( = 3\sqrt 3 \)

\( \Leftrightarrow \dfrac{{\sin x}}{{\cos x}}\)\( + \left[ {\dfrac{{\sin \left( {x + \dfrac{\pi }{3}} \right)}}{{\cos \left( {x + \dfrac{\pi }{3}} \right)}} + \dfrac{{\sin \left( {x + \dfrac{{2\pi }}{3}} \right)}}{{\cos \left( {x + \dfrac{{2\pi }}{3}} \right)}}} \right] = 3\sqrt 3 \)

\( \Leftrightarrow \dfrac{{\sin x}}{{\cos x}}\)\( + \dfrac{{\sin \left( {x + \dfrac{\pi }{3}} \right).\cos \left( {x + \dfrac{{2\pi }}{3}} \right) + \cos \left( {x + \dfrac{\pi }{3}} \right).\sin \left( {x + \dfrac{{2\pi }}{3}} \right)}}{{\cos \left( {x + \dfrac{\pi }{3}} \right)\cos \left( {x + \dfrac{{2\pi }}{3}} \right)}}\)\( = 3\sqrt 3 \)

\( \Leftrightarrow \dfrac{{\sin x}}{{\cos x}} + \dfrac{{\sin \left[ {\left( {x + \dfrac{\pi }{3}} \right) + \left( {x + \dfrac{{2\pi }}{3}} \right)} \right]}}{{\cos \left( {x + \dfrac{\pi }{3}} \right).\cos \left( {x + \dfrac{{2\pi }}{3}} \right)}} = 3\sqrt 3 {\rm{ }}\)

$ \Leftrightarrow \dfrac{{\sin x}}{{\cos x}} + \dfrac{{\sin \left( {2x + \pi } \right)}}{{\cos \left( {x + \dfrac{\pi }{3}} \right)\cos \left( {x + \dfrac{{2\pi }}{3}} \right)}}$$ = 3\sqrt 3 $
$ \Leftrightarrow \dfrac{{\sin x}}{{\cos x}} - \dfrac{{\sin 2x}}{{\dfrac{1}{2}\left[ {\cos \left( {x + \dfrac{{2\pi }}{3} + x + \dfrac{\pi }{3}} \right) + \cos \left( {x + \dfrac{{2\pi }}{3} - x - \dfrac{\pi }{3}} \right)} \right]}} $$= 3\sqrt 3 $
$ \Leftrightarrow \dfrac{{\sin x}}{{\cos x}} - \dfrac{{2\sin 2x}}{{\cos \left( {2x + \pi } \right) + \cos \left( {\dfrac{\pi }{3}} \right)}} = 3\sqrt 3 $
$\begin{array}{l}
\Leftrightarrow \dfrac{{\sin x}}{{\cos x}} - \dfrac{{2\sin 2x}}{{ - \cos 2x + \dfrac{1}{2}}} = 3\sqrt 3 \\
\Leftrightarrow \dfrac{{\sin x}}{{\cos x}} - \dfrac{{2.2\sin 2x}}{{ - 2\cos 2x + 2.\dfrac{1}{2}}} = 3\sqrt 3
\end{array}$
$ \Leftrightarrow \dfrac{{\sin x}}{{\cos x}} - \dfrac{{4\sin 2x}}{{1 - 2\cos 2x}} = 3\sqrt 3 $
$ \Leftrightarrow \dfrac{{\sin x\left( {1 - 2\cos 2x} \right) - 4\sin 2x.\cos x}}{{\cos x\left( {1 - 2\cos 2x} \right)}} = 3\sqrt 3 $
$ \Leftrightarrow \dfrac{{\sin x - 2\sin x\cos 2x - 4\sin 2x\cos x}}{{\cos x\left( {1 - 2\cos 2x} \right)}} = 3\sqrt 3 $
$ \Leftrightarrow \dfrac{{\sin x - 2.\dfrac{1}{2}\left( {\sin 3x - \sin x} \right) - 4.\dfrac{1}{2}\left( {\sin 3x + \sin x} \right)}}{{\cos x\left( {1 - 2\cos 2x} \right)}} = 3\sqrt 3 $
$ \Leftrightarrow \dfrac{{\sin x - \sin 3x + \sin x - 2\sin 3x - 2\sin x}}{{\cos x - \cos x - \cos 3x}} = 3\sqrt 3 $
$ \Leftrightarrow \dfrac{{ - 3\sin 3x}}{{ - \cos 3x}} = 3\sqrt 3 $
$ \Leftrightarrow 3\tan 3x = 3\sqrt 3 \Leftrightarrow \tan 3x = \sqrt 3 $
\( \Leftrightarrow 3x = \dfrac{\pi }{3} + k\pi \Leftrightarrow x = \dfrac{\pi }{9} + \dfrac{{k\pi }}{3}\).
Kiểm tra ta thấy nghiệm \(x = \dfrac{\pi }{9} + \dfrac{{k\pi }}{3}\) thỏa mãn các điều kiện của phương trình đầu.
Do đó phương trình \(\tan 3x = \sqrt 3 \) tương đương với phương trình ban đầu (có cùng tập nghiệm).