To prove the identity \(\cos 4\theta + 4\cos 2\theta + 3 \equiv 8\cos^4 \theta\), we start by expressing \(\cos 4\theta\) and \(\cos 2\theta\) in terms of \(\cos \theta\).

First, use the double angle formula for cosine:

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

Next, express \(\cos 4\theta\) using the double angle formula again:

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

Substitute \(\cos 2\theta = 2\cos^2 \theta - 1\) into the expression for \(\cos 4\theta\):

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

Expand \((2\cos^2 \theta - 1)^2\):

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

Substitute back into the expression for \(\cos 4\theta\):

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

Now substitute \(\cos 4\theta\) and \(\cos 2\theta\) into the original identity:

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

Simplify the expression:

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

Thus, the identity is proven: \(\cos 4\theta + 4\cos 2\theta + 3 \equiv 8\cos^4 \theta\).