Part (i):

We need to prove the identity \(\cos 4\theta + 4\cos 2\theta \equiv 8\cos^4 \theta - 3\).

Using the double angle formulas:

\(\cos 2\theta = 2\cos^2 \theta - 1\)

\(\cos 4\theta = 2\cos^2 2\theta - 1 = 2(2\cos^2 \theta - 1)^2 - 1\)

Expanding \(\cos 4\theta\):

\(\cos 4\theta = 2(4\cos^4 \theta - 4\cos^2 \theta + 1) - 1\)

\(= 8\cos^4 \theta - 8\cos^2 \theta + 2 - 1\)

\(= 8\cos^4 \theta - 8\cos^2 \theta + 1\)

Now, substitute \(\cos 4\theta\) and \(\cos 2\theta\) into the left-hand side:

\(\cos 4\theta + 4\cos 2\theta = (8\cos^4 \theta - 8\cos^2 \theta + 1) + 4(2\cos^2 \theta - 1)\)

\(= 8\cos^4 \theta - 8\cos^2 \theta + 1 + 8\cos^2 \theta - 4\)

\(= 8\cos^4 \theta - 3\)

Thus, the identity is proven.

Part (ii):

We need to solve \(\cos 4\theta + 4\cos 2\theta = 2\).

Using the identity from part (i):

\(8\cos^4 \theta - 3 = 2\)

\(8\cos^4 \theta = 5\)

\(\cos^4 \theta = \frac{5}{8}\)

\(\cos^2 \theta = \sqrt{\frac{5}{8}}\)

\(\cos \theta = \pm \sqrt{\frac{5}{8}}\)

Calculating \(\theta\) for \(\cos \theta = \sqrt{\frac{5}{8}}\):

\(\theta = \cos^{-1}(\sqrt{\frac{5}{8}}) \approx 27.3^\circ\)

Other solutions in the range \(0^\circ \leq \theta \leq 360^\circ\) are:

\(360^\circ - 27.3^\circ = 332.8^\circ\)

For \(\cos \theta = -\sqrt{\frac{5}{8}}\):

\(\theta = 180^\circ - 27.3^\circ = 152.8^\circ\)

\(180^\circ + 27.3^\circ = 207.2^\circ\)

Thus, \(\theta = 27.3^\circ, 152.8^\circ, 207.2^\circ, 332.8^\circ\).