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

Substitute this into the equation:

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

Simplify the equation:

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

\(\sin \theta = 5 - 6\sin^2 \theta\)

Rearrange to form a quadratic equation:

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

Let \(x = \sin \theta\). The equation becomes:

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

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

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

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

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

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

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

For \(x = \frac{5}{6}\), \(\theta = \sin^{-1}\left(\frac{5}{6}\right) \approx 56.4^\circ\) and \(180^\circ - 56.4^\circ = 123.6^\circ\).

For \(x = -1\), \(\theta = 270^\circ\).

Thus, the solutions are \(\theta = 56.4^\circ, 123.6^\circ, 270^\circ\).