18. Find the sum of all real roots of the equation $3 \tan ^{2} x+8 \tan x+3=0$ in the range $0<x<2 \pi$.求方程 $3 \tan ^{2} x+8 \tan x+3=0$ 在區間 $0<x<2 \pi$ 內所有實根之和。
18. $5 \pi$
18. Being a quadratic equation in $\tan x$ with discriminant $8^{2}-4(3)(3)>0$, we see that there are two possible values of $\tan x$ (say $\tan x_{1}$ and $\tan x_{2}$ ) and hence four possible values of $x$ in the range $0<x<2 \pi$. Since $\tan x_{1}+\tan x_{2}=-\frac{8}{3}$ and $\tan x_{1} \tan x_{2}=\frac{3}{3}=1$, both $\tan x_{1}$ and $\tan x_{2}$ are negative, so we have two possible values of $x$ in the second quadrant and two in the fourth quadrant.
Consider $x_{1}, x_{2}$ in the second quadrant satisfying the equation. From $\tan x_{1} \tan x_{2}=1$, we get $\tan x_{2}=\cot x_{1}=\tan \left(\frac{3 \pi}{2}-x_{1}\right)$. As $\frac{3 \pi}{2}-x_{1}$ is also in the second quadrant, we must have $x_{2}=\frac{3 \pi}{2}-x_{1}$. On the other hand, it is easy to see that the two possible values of $x$ in the fourth quadrant are simply $x_{1}+\pi$ and $x_{2}+\pi$, as $\tan (x+\pi)=\tan x$ for all $x$ (except at points where $\tan x$ is undefined). It follows that the answer is
$$
x_{1}+\left(\frac{3 \pi}{2}-x_{1}\right)+\left(x_{1}+\pi\right)+\left(\frac{3 \pi}{2}-x_{1}+\pi\right)=5 \pi \text {. }
$$