Answer: \(x=0,\frac{\pi}{3},\frac{\pi}{2},\frac{5\pi}{6}\).

The key point is to reduce the equation to a standard trigonometric equation, then use the given interval to list all valid solutions.

Since

\(\operatorname{cot}^2\left(2x-\frac{\pi}{3}\right)=\frac13,\)

we have

\(\tan^2\left(2x-\frac{\pi}{3}\right)=3.\)

Therefore

\(\tan\left(2x-\frac{\pi}{3}\right)=\pm\sqrt3.\)

Let

\(u=2x-\frac{\pi}{3}.\)

Since \(0\leq x\lt \pi\),

\(-\frac{\pi}{3}\leq u\lt \frac{5\pi}{3}.\)

In this interval, \(\tan u=\pm\sqrt3\) when

So

Adding \(\frac{\pi}{3}\) and dividing by \(2\),