Answer
The solution set to this equation is $$\{\frac{\pi}{18},\frac{7\pi}{18},\frac{13\pi}{18},\frac{19\pi}{18},\frac{25\pi}{18},\frac{31\pi}{18}\}$$
Work Step by Step
$$3\tan3x=\sqrt3$$ over interval $[0,2\pi)$
1) Interval $[0,2\pi)$ can be written as
$$0\le x\lt2\pi$$
That means, for $3x$, the interval would be
$$0\le3x\lt6\pi$$
or $$3x\in[0,6\pi)$$
2) Now consider back the equation $$3\tan3x=\sqrt3$$
$$\tan3x=\frac{\sqrt3}{3}$$
Over the interval $[0,6\pi)$, there are 6 values whose $\tan$ equals $\frac{\sqrt3}{3}$, which are $\frac{\pi}{6},\frac{7\pi}{6},\frac{13\pi}{6},\frac{19\pi}{6},\frac{25\pi}{6},\frac{31\pi}{6}$, meaning that
$$3x=\{\frac{\pi}{6},\frac{7\pi}{6},\frac{13\pi}{6},\frac{19\pi}{6},\frac{25\pi}{6},\frac{31\pi}{6}\}$$
So $$x=\{\frac{\pi}{18},\frac{7\pi}{18},\frac{13\pi}{18},\frac{19\pi}{18},\frac{25\pi}{18},\frac{31\pi}{18}\}$$
