Answer: (a) \(x=-165^\circ,-75^\circ,15^\circ,105^\circ\). (b) \(-\frac{\pi}{6}\lt \theta\lt \frac{\pi}{6}\), \(\theta\ne0\), or \(\frac{5\pi}{6}\lt \theta\lt \frac{7\pi}{6}\), \(\theta\ne\pi\).

Use the standard trigonometric identities and make sure the final angles are chosen from the interval given in the question.

The arithmetic progression has first term \(2\tan2x\) and common difference \(3\tan2x\).

Using \(S_n=\frac n2(2a+(n-1)d)\),

\(S_{30}=15(4\tan2x+29\cdot3\tan2x)=1365\tan2x\).

Since \(S_{30}=455\sqrt3\),

\(\tan2x=\frac{455\sqrt3}{1365}=\frac{\sqrt3}{3}\).

For \(-180^\circ\leq x\leq180^\circ\), this gives

\(x=-165^\circ,-75^\circ,15^\circ,105^\circ\).

For the geometric progression, the common ratio is

\(r=\frac{20\cos^4(\theta-\frac{\pi}{2})}{5\cos^2(\theta-\frac{\pi}{2})}=4\cos^2\left(\theta-\frac{\pi}{2}\right)\).

A sum to infinity requires \(|r|\lt 1\). Since \(r\geq0\),

\(4\cos^2\left(\theta-\frac{\pi}{2}\right)\lt 1\).

The critical values in the given interval are

\(\theta=-\frac{\pi}{6},\frac{\pi}{6},\frac{5\pi}{6},\frac{7\pi}{6}\).

Therefore

\(-\frac{\pi}{6}\lt \theta\lt \frac{\pi}{6}\), excluding \(\theta=0\), or

\(\frac{5\pi}{6}\lt \theta\lt \frac{7\pi}{6}\), excluding \(\theta=\pi\).

This gives the required answer: (a) \(x=-165^\circ,-75^\circ,15^\circ,105^\circ\). (b) \(-\frac{\pi}{6}\lt \theta\lt \frac{\pi}{6}\), \(\theta\ne0\), or \(\frac{5\pi}{6}\lt \theta\lt \frac{7\pi}{6}\), \(\theta\ne\pi\).