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

Substitute this into the equation:

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

Simplify and rearrange:

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

\(6\cos^2 \theta - 3\cos \theta - 5 = 0\)

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

\(6x^2 - 3x - 5 = 0\)

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 6\), \(b = -3\), \(c = -5\):

\(x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 6 \cdot (-5)}}{2 \cdot 6}\)

\(x = \frac{3 \pm \sqrt{9 + 120}}{12}\)

\(x = \frac{3 \pm \sqrt{129}}{12}\)

Calculate the values:

\(x_1 = \frac{3 + \sqrt{129}}{12} \approx 0.694\)

\(x_2 = \frac{3 - \sqrt{129}}{12} \approx -1.194\)

Since \(\cos \theta\) must be between -1 and 1, only \(x_1\) is valid.

Find \(\theta\) using \(\cos^{-1}(x_1)\):

\(\theta = \cos^{-1}(0.694) \approx 134.1^\circ\)

The second solution in the range \([0^\circ, 360^\circ]\) is:

\(\theta = 360^\circ - 134.1^\circ = 225.9^\circ\)

Thus, the solutions are \(\theta = 134.1^\circ\) and \(\theta = 225.9^\circ\).