(i) Start with the equation \(3 \cos^4 \theta + 4 \sin^2 \theta - 3 = 0\).

Use the identity \(\sin^2 \theta = 1 - \cos^2 \theta\).

Substitute \(\sin^2 \theta = 1 - x\) where \(x = \cos^2 \theta\).

The equation becomes \(3x^2 + 4(1-x) - 3 = 0\).

Simplify to get \(3x^2 - 4x + 1 = 0\).

(ii) Solve the quadratic equation \(3x^2 - 4x + 1 = 0\).

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

Calculate the discriminant: \((-4)^2 - 4 \times 3 \times 1 = 4\).

\(x = \frac{4 \pm 2}{6}\), giving \(x = 1\) or \(x = \frac{1}{3}\).

For \(x = 1\), \(\cos^2 \theta = 1\), so \(\cos \theta = \pm 1\).

\(\theta = 0^\circ, 180^\circ\).

For \(x = \frac{1}{3}\), \(\cos^2 \theta = \frac{1}{3}\), so \(\cos \theta = \pm \frac{1}{\sqrt{3}}\).

\(\theta = 54.7^\circ, 125.3^\circ\).