Điều kiện xác định:

$\left\{ \begin{array}{l} \cos x \ne 0\\ \cos 3x \ne 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x \ne \dfrac{\pi }{2} + k\pi \\ 3x \ne \dfrac{\pi }{2} + k\pi  \end{array} \right. \Rightarrow \left\{ \begin{array}{l} x \ne \dfrac{\pi }{2} + k\pi \\ x \ne \dfrac{\pi }{6} + \dfrac{{k\pi }}{3} \end{array} \right.\left( {k \in Z} \right)(1)$

$\begin{array}{l}
\tan x = \tan 3x\\
 \Leftrightarrow 3x = x + k\pi \\
 \Leftrightarrow x = \dfrac{{k\pi }}{2}\\
\left( 1 \right) \Rightarrow x = k\pi \\
0 < x < 30 \Rightarrow 0 < k\pi  < 30 \Rightarrow 0 < k < \dfrac{{30}}{\pi }\\
 \Rightarrow k \in \left[ {1;9} \right]\\
 \Rightarrow S = \pi  + 2\pi  + 3\pi  + ... + 9\pi  = \pi \left( {1 + 2 + 3 + ... + 9} \right)\\
 = \pi .\dfrac{{9.10}}{2} = 45\pi 
\end{array}$

`~rai~`

\(\tan x=\tan3x\quad(1)\\ĐKXĐ:\begin{cases}\cos x\ne 0\\\cos3x\ne 0\end{cases}\\\Leftrightarrow \begin{cases}x\ne\dfrac{\pi}{2}+k\pi\\3x\ne \dfrac{\pi}{2}+k\pi\end{cases}\\\Leftrightarrow \begin{cases}x\ne\dfrac{\pi}{2}+k\pi\\x\ne\dfrac{\pi}{6}+k\dfrac{\pi}{3}\end{cases}\\\Leftrightarrow x\ne\dfrac{\pi}{6}+k\dfrac{\pi}{3}.(k\in\mathbb{Z})\\(1)\Leftrightarrow 3x=x+k\pi\\\Leftrightarrow 2x=k\pi\\\Leftrightarrow x=k\dfrac{\pi}{2}.(k\in\mathbb{Z})\\\text{Kết hợp nghiệm và ĐKXĐ trên đường tròn lượng giác được:}\\x=k\pi.(k\in\mathbb{Z})\\\text{Do }x\in(0;30)\\\Leftrightarrow 0<x<30\\\Leftrightarrow 0<k\pi<30\\\Leftrightarrow 0<k<\dfrac{30}{\pi}\\\text{mà }k\in\mathbb{Z}\Rightarrow k\in\{1;2;3;\cdots;9\}\\\Rightarrow x\in\{\pi;2\pi;3\pi;\cdots;9\pi\}.\\\text{Tổng các nghiệm:}S=\pi+2\pi+3\pi+\cdots+9\pi=45\pi.\)