To solve the equation \(\cos \theta + 4 \cos 2\theta = 3\), we start by using the double angle identity \(\cos 2\theta = 2\cos^2\theta - 1\).

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

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

Simplify the equation:

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

Rearrange terms:

\(8\cos^2\theta + \cos \theta - 7 = 0\)

This is a quadratic equation in \(\cos \theta\). Let \(x = \cos \theta\), then:

\(8x^2 + x - 7 = 0\)

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 8\), \(b = 1\), \(c = -7\):

\(x = \frac{-1 \pm \sqrt{1^2 - 4 \times 8 \times (-7)}}{2 \times 8}\)

\(x = \frac{-1 \pm \sqrt{1 + 224}}{16}\)

\(x = \frac{-1 \pm \sqrt{225}}{16}\)

\(x = \frac{-1 \pm 15}{16}\)

So, \(x = \frac{14}{16} = \frac{7}{8}\) or \(x = \frac{-16}{16} = -1\).

Thus, \(\cos \theta = \frac{7}{8}\) or \(\cos \theta = -1\).

For \(\cos \theta = \frac{7}{8}\):

\(\theta = \cos^{-1}\left(\frac{7}{8}\right) \approx 29^\circ\)

For \(\cos \theta = -1\):

\(\theta = 180^\circ\)

Therefore, the solutions are \(\theta = 29^\circ\) and \(\theta = 180^\circ\).