(i) Start with the given equation:

\(4 \sin^4 \theta + 5 = 7 \cos^2 \theta\)

Using the identity \(\sin^2 \theta + \cos^2 \theta = 1\), we have \(\cos^2 \theta = 1 - \sin^2 \theta\).

Substitute \(\cos^2 \theta\) in the equation:

\(4 \sin^4 \theta + 5 = 7(1 - \sin^2 \theta)\)

\(4 \sin^4 \theta + 5 = 7 - 7 \sin^2 \theta\)

Let \(x = \sin^2 \theta\), then \(\sin^4 \theta = x^2\).

Substitute \(x\) into the equation:

\(4x^2 + 5 = 7 - 7x\)

Rearrange to form a quadratic equation:

\(4x^2 + 7x - 2 = 0\)

(ii) Solve the quadratic equation \(4x^2 + 7x - 2 = 0\).

Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 4\), \(b = 7\), \(c = -2\):

\(x = \frac{-7 \pm \sqrt{7^2 - 4 \cdot 4 \cdot (-2)}}{2 \cdot 4}\)

\(x = \frac{-7 \pm \sqrt{49 + 32}}{8}\)

\(x = \frac{-7 \pm \sqrt{81}}{8}\)

\(x = \frac{-7 \pm 9}{8}\)

\(x = \frac{2}{8} = \frac{1}{4} \quad \text{or} \quad x = \frac{-16}{8} = -2\)

Since \(x = \sin^2 \theta\), \(x = -2\) is not valid as \(\sin^2 \theta\) must be between 0 and 1.

Thus, \(\sin^2 \theta = \frac{1}{4}\).

\(\sin \theta = \pm \frac{1}{2}\).

For \(\sin \theta = \frac{1}{2}\), \(\theta = 30^\circ, 150^\circ\).

For \(\sin \theta = -\frac{1}{2}\), \(\theta = 210^\circ, 330^\circ\).

Therefore, \(\theta = 30^\circ, 150^\circ, 210^\circ, 330^\circ\).