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

Reduce the equation to a standard trigonometric form, find the principal solutions, then add the correct multiples of the period to cover the required interval.

Let

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

The equation becomes

\(3\operatorname{cosec}^2u=4.\)

So

\(\operatorname{cosec}^2u=\frac43,\)

and therefore

\(\sin^2u=\frac34.\)

Hence

\(\sin u=\pm\frac{\sqrt3}{2}.\)

Since \(0\lt x\leq3\pi\),

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

In this interval,

\(u=\frac{\pi}{3},\ \frac{2\pi}{3},\ \frac{4\pi}{3},\ \frac{5\pi}{3}.\)

Now

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

so

\(x=\frac32\left(u+\frac{\pi}{3}\right).\)

This gives

\(x=\pi,\ \frac{3\pi}{2},\ \frac{5\pi}{2},\ 3\pi.\)