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

First, use the double angle formula for \(\cos 2\theta\):

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

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:

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

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

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

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

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

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

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

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