Start with the equation \(3 \sin^2 \theta = 4 \cos \theta - 1\).

Use the identity \(\sin^2 \theta = 1 - \cos^2 \theta\) to rewrite the equation:

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

Expand and simplify:

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

Rearrange to form a quadratic equation in \(\cos \theta\):

\(3 \cos^2 \theta + 4 \cos \theta - 4 = 0\)

Let \(c = \cos \theta\). The equation becomes:

\(3c^2 + 4c - 4 = 0\)

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

\(c = \frac{-4 \pm \sqrt{4^2 - 4 \times 3 \times (-4)}}{2 \times 3}\)

\(c = \frac{-4 \pm \sqrt{16 + 48}}{6}\)

\(c = \frac{-4 \pm \sqrt{64}}{6}\)

\(c = \frac{-4 \pm 8}{6}\)

Solutions for \(c\) are \(c = \frac{2}{3}\) or \(c = -2\).

Since \(\cos \theta\) must be between -1 and 1, \(c = -2\) is not valid.

Thus, \(\cos \theta = \frac{2}{3}\).

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

\(\theta = \cos^{-1}\left(\frac{2}{3}\right) \approx 48.2^\circ\)

The second solution in the range \(0^\circ \leq \theta \leq 360^\circ\) is:

\(\theta = 360^\circ - 48.2^\circ = 311.8^\circ\)

Therefore, \(\theta = 48.2^\circ\) or \(\theta = 311.8^\circ\).