For a geometric progression to be convergent, the common ratio \(r\) must satisfy \(-1 < r < 1\).

The common ratio \(r\) is given by:

\(r = \frac{2 \cos \theta}{\sqrt{3}}\)

Thus, the inequality becomes:

\(-1 < \frac{2 \cos \theta}{\sqrt{3}} < 1\)

We solve the inequality \(0 < \frac{2 \cos \theta}{\sqrt{3}} < 1\) since \(0 < \theta < \pi\):

\(0 < 2 \cos \theta < \sqrt{3}\)

\(0 < \cos \theta < \frac{\sqrt{3}}{2}\)

The values of \(\theta\) for which \(\cos \theta = \frac{\sqrt{3}}{2}\) are \(\theta = \frac{\pi}{6}\) and \(\theta = \frac{5\pi}{6}\).

Therefore, the set of values of \(\theta\) for which the progression is convergent is:

\(\frac{\pi}{6} < \theta < \frac{5\pi}{6}\).