(a) Start by expressing \(\cos 4\theta\) and \(\cos 2\theta\) in terms of \(\cos \theta\):

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

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

Substitute \(\cos 2\theta\) into \(\cos 4\theta\):

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

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

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

Now, add \(4\cos 2\theta + 3\):

\(\cos 4\theta + 4\cos 2\theta + 3 = 8\cos^4 \theta - 8\cos^2 \theta + 1 + 4(2\cos^2 \theta - 1) + 3\)

\(= 8\cos^4 \theta - 8\cos^2 \theta + 1 + 8\cos^2 \theta - 4 + 3\)

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

Thus, the identity is proven.

(b) Using the identity, solve:

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

\(8\cos^4 \theta = 7\)

\(\cos^4 \theta = \frac{7}{8}\)

\(\cos \theta = \pm \sqrt[4]{\frac{7}{8}}\)

Calculate \(\theta\) for \(0^\circ \leq \theta \leq 180^\circ\):

\(\theta = \cos^{-1}(\sqrt[4]{\frac{7}{8}}) \approx 14.7^\circ\)

\(\theta = 180^\circ - 14.7^\circ = 165.3^\circ\)

Thus, \(\theta = 14.7^\circ, 165.3^\circ\).