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

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

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

Simplify the equation:

\(\cos \theta + 6\cos^2 \theta - 3 = 2\)

Rearrange terms:

\(6\cos^2 \theta + \cos \theta - 5 = 0\)

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

\(6x^2 + x - 5 = 0\)

Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 6\), \(b = 1\), \(c = -5\):

\(x = \frac{-1 \pm \sqrt{1^2 - 4 \times 6 \times (-5)}}{2 \times 6}\)

\(x = \frac{-1 \pm \sqrt{1 + 120}}{12}\)

\(x = \frac{-1 \pm \sqrt{121}}{12}\)

\(x = \frac{-1 \pm 11}{12}\)

\(x = \frac{10}{12} = \frac{5}{6} \quad \text{or} \quad x = \frac{-12}{12} = -1\)

Thus, \(\cos \theta = \frac{5}{6}\) or \(\cos \theta = -1\).

For \(\cos \theta = \frac{5}{6}\):

\(\theta = \cos^{-1} \left( \frac{5}{6} \right) \approx 33.6^\circ\)

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

\(\theta = 180^\circ\)

Therefore, the solutions are \(\theta = 33.6^\circ\) and \(\theta = 180^\circ\).