#### Answer

The solution set is $$\{\frac{\pi}{6}, \frac{2\pi}{3}\}$$

#### Work Step by Step

$$3\cot2x-\sqrt3=0$$
over the interval $[0,2\pi)$
We carry out algebraic operations here normally as any equations:
$$\cot2x=\frac{\sqrt3}{3}$$
Now we can solve for $2x$.
Over the interval $[0,2\pi)$, there are 2 values of $2x$ with which $\cot2x=\frac{\sqrt3}{3}$, which are $\frac{\pi}{3}$ and $\frac{4\pi}{3}$.
Therefore, $$2x=\{\frac{\pi}{3},\frac{4\pi}{3}\}$$
That means, $$x=\{\frac{\pi}{6}, \frac{2\pi}{3}\}$$
This is the solution set of the problem.