Start with the equation:

\(3 \tan^2 \theta + 1 = \frac{2}{\tan^2 \theta}\)

Multiply through by \(\tan^2 \theta\) to eliminate the fraction:

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

Let \(x = \tan^2 \theta\), then the equation becomes:

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

Factor the quadratic equation:

\((3x - 2)(x + 1) = 0\)

Thus, \(x = \frac{2}{3}\) or \(x = -1\).

Since \(x = \tan^2 \theta\), \(x\) must be non-negative, so \(x = -1\) is not valid.

Therefore, \(\tan^2 \theta = \frac{2}{3}\).

Take the square root to find \(\tan \theta\):

\(\tan \theta = \pm \sqrt{\frac{2}{3}}\)

Calculate \(\theta\) using the inverse tangent function:

\(\theta = \tan^{-1}(\sqrt{\frac{2}{3}}) \approx 39.2^\circ\)

Since \(\tan \theta\) is positive in the first and third quadrants, the solutions are:

\(\theta = 39.2^\circ\) and \(\theta = 180^\circ - 39.2^\circ = 140.8^\circ\)