Điều kiện: \(\left\{ \begin{array}{l}x \ne \frac{\pi }{2} + k\pi \\x \ne \frac{\pi }{4} + k\frac{\pi }{2}\end{array} \right.,k \in \mathbb{Z}.\)

Ta có: \(\frac{{\sin x}}{{\cos x}} + \frac{{\sin 2x}}{{\cos 2x}} = \frac{{\sin x\cos 2x + \cos x\sin 2x}}{{{\mathop{\rm cosx}\nolimits} .cos2x}} = \frac{{\sin 3x}}{{\cos 2x.\cos x}}\)

Suy ra:\(\tan x + \tan 2x = \sin 3x.\cos x \Leftrightarrow \frac{{\sin 3x}}{{\cos 2x.\cos x}} = \sin 3x.\cos x\)

\(\begin{array}{l} \Leftrightarrow \sin 3x = \sin 3x.\cos 2x.co{s^2}x\\ \Leftrightarrow \sin 3x(\cos 2x.{\cos ^2}x - 1) = 0\\ \Leftrightarrow (2{\cos ^4}x - {\cos ^2}x - 1)\sin 3x = 0\\ \Leftrightarrow \left[ \begin{array}{l}\sin 3x = 0\,(1)\\2{\cos ^4}x - {\cos ^2}x - 1\, = 0(2)\end{array} \right.\end{array}\)

Giải (1): \(\sin 3x = 0 \Leftrightarrow x = \frac{{k\pi }}{3},k \in \mathbb{Z}\,(*)\)

Giải (2): Đặt \(t = {\cos ^2}x,0 \le t \le 1,\) Bất phương trình trở thành:

\(2{t^4} - {t^2} - 1 = 0 \Leftrightarrow \left[ \begin{array}{l}t = 1\\t =  - \frac{1}{2}\,(loai)\end{array} \right.\)

Với \(t = 1 \Rightarrow {\cos ^2}x = 1 \Leftrightarrow x = k\pi ,k \in \mathbb{Z}\,(**)\)

Từ \((*);(**) \Rightarrow x = \frac{{k\pi }}{3},k \in \mathbb{Z}\)