To find the maximum of \(f(\theta) = \frac{4}{6 + 2\cos \theta}\) for \(\frac{\pi}{2} \leq \theta \leq 2\pi\), simplify to \(f(\theta) = \frac{2}{3 + \cos \theta}\). The maximum occurs when \(3 + \cos \theta\) is minimized. In the interval, \(\cos \theta\) ranges from \(-1\) to \(1\), with minimum \(-1\) at \(\theta = \pi\). Thus, minimum denominator is \(3 - 1 = 2\), so maximum \(f(\theta) = \frac{2}{2} = 1\).