(i) Start with the equation \(3 \sin \theta = \cos \theta\). Divide both sides by \(\cos \theta\) to get \(3 \tan \theta = 1\). Therefore, \(\tan \theta = \frac{1}{3}\). Solving for \(\theta\) in the range \(0^\circ < \theta < 180^\circ\), we find \(\theta = 18.4^\circ\).

(ii) Given \(3 \sin^2 2x = \cos^2 2x\), rewrite it as \(\frac{\sin^2 2x}{\cos^2 2x} = \frac{1}{3}\). This implies \(\tan^2 2x = \frac{1}{3}\), so \(\tan 2x = \pm \frac{1}{\sqrt{3}}\). The solutions for \(2x\) are \(30^\circ, 150^\circ, 210^\circ, 330^\circ\). Dividing by 2, we get \(x = 15^\circ, 75^\circ, 105^\circ, 165^\circ\).