Start with the equation:

\(3 \sin^2 2\theta + 8 \cos 2\theta = 0\)

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

\(3(1 - \cos^2 2\theta) + 8 \cos 2\theta = 0\)

Simplify to:

\(3 - 3\cos^2 2\theta + 8 \cos 2\theta = 0\)

Rearrange to form a quadratic equation in \(\cos 2\theta\):

\(3\cos^2 2\theta - 8 \cos 2\theta - 3 = 0\)

Let \(x = \cos 2\theta\), then solve the quadratic:

\(3x^2 - 8x - 3 = 0\)

Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = -8\), \(c = -3\):

\(x = \frac{8 \pm \sqrt{(-8)^2 - 4 \times 3 \times (-3)}}{2 \times 3}\)

\(x = \frac{8 \pm \sqrt{64 + 36}}{6}\)

\(x = \frac{8 \pm \sqrt{100}}{6}\)

\(x = \frac{8 \pm 10}{6}\)

\(x = 3.0 \text{ or } x = -\frac{1}{3}\)

Since \(\cos 2\theta\) must be between -1 and 1, we take \(\cos 2\theta = -\frac{1}{3}\).

Find \(2\theta\):

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

\(2\theta \approx 109.47^\circ \text{ or } 250.53^\circ\)

Thus, \(\theta \approx 54.7^\circ \text{ or } 125.3^\circ\).