Answer
$x\displaystyle \in\{\frac{\pi}{18},\frac{7\pi}{18},\frac{13\pi}{18},\frac{19\pi}{18}, \frac{25\pi}{18},\frac{31\pi}{18}\}$
Work Step by Step
If $ 0 \leq x < 2\pi$ then
$ 0 \leq 3x < 6\pi$
So, $3x\in[0,6\pi)$, meaning that
if 3x (a solution) is any number from $[0,2\pi)$, then
$(3x+2\pi)$ and $(3x+4\pi)$ are also solutions.
Divide the equation with 3.
$\displaystyle \tan 3x=\frac{\sqrt{3}}{3}$
In $[0,2\pi),\ \tan \displaystyle \frac{\pi}{6}=\frac{\sqrt{3}}{3}, \tan \displaystyle \frac{7\pi}{6}=\frac{\sqrt{3}}{3}$
so, $3x$ can be
$\displaystyle \frac{\pi}{6} \qquad $or $\displaystyle \qquad \frac{7\pi}{6}.$
3x can also be
$\displaystyle \frac{\pi}{6}+2\pi=\frac{13\pi}{6} \quad $or $\displaystyle \quad \frac{7\pi}{6}+2\pi=\frac{19\pi}{6}$
And, also
$\displaystyle \frac{\pi}{6}+4\pi=\frac{25\pi}{6} \quad $or $\displaystyle \quad \frac{7\pi}{6}+4\pi=\frac{31\pi}{6}$
So,
$3x\displaystyle \in\{\frac{\pi}{6},\frac{7\pi}{6},\frac{13\pi}{6},\frac{19\pi}{6}, \frac{25\pi}{6},\frac{31\pi}{6}\}$,
and, since $(3x)\div 3=x,$
after dividing each corresponding angle with 3,
$x\displaystyle \in\{\frac{\pi}{18},\frac{7\pi}{18},\frac{13\pi}{18},\frac{19\pi}{18}, \frac{25\pi}{18},\frac{31\pi}{18}\}$
