(i) Start with the equation \(\sin^2 \theta + 3 \sin \theta \cos \theta = 4 \cos^2 \theta\).

Divide the entire equation by \(\cos^2 \theta\):

\(\frac{\sin^2 \theta}{\cos^2 \theta} + 3 \frac{\sin \theta \cos \theta}{\cos^2 \theta} = 4\).

This simplifies to \(\tan^2 \theta + 3 \tan \theta = 4\).

Thus, the equation can be written as a quadratic in \(\tan \theta\):

\(\tan^2 \theta + 3 \tan \theta - 4 = 0\).

(ii) Solve the quadratic equation \(\tan^2 \theta + 3 \tan \theta - 4 = 0\).

Using the quadratic formula \(\tan \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 3\), \(c = -4\):

\(\tan \theta = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1}\).

\(\tan \theta = \frac{-3 \pm \sqrt{9 + 16}}{2}\).

\(\tan \theta = \frac{-3 \pm 5}{2}\).

This gives \(\tan \theta = 1\) or \(\tan \theta = -4\).

For \(\tan \theta = 1\), \(\theta = 45^\circ\).

For \(\tan \theta = -4\), \(\theta = 104^\circ\).

Thus, \(\theta = 45^\circ\) or \(\theta = 104^\circ\).