To express \(\cot 2\theta\) in terms of \(\tan \theta\), use the identity:

\(\cot 2\theta = \frac{1 - \tan^2 \theta}{2\tan \theta}\)

Given the equation:

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

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

\(1 - \tan^2 \theta = 2\tan \theta + 2\tan^2 \theta\)

Rearrange to form a quadratic equation:

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

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

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

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 3\), \(b = 2\), \(c = -1\):

\(x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3}\)

\(x = \frac{-2 \pm \sqrt{4 + 12}}{6}\)

\(x = \frac{-2 \pm \sqrt{16}}{6}\)

\(x = \frac{-2 \pm 4}{6}\)

\(x = \frac{2}{6} = \frac{1}{3} \quad \text{or} \quad x = \frac{-6}{6} = -1\)

Thus, \(\tan \theta = \frac{1}{3}\) or \(\tan \theta = -1\).

For \(\tan \theta = \frac{1}{3}\), \(\theta = \tan^{-1}\left(\frac{1}{3}\right) \approx 18.4^\circ\).

For \(\tan \theta = -1\), \(\theta = 135^\circ\) (since \(\tan 135^\circ = -1\)).

Thus, the solutions are \(\theta = 18.4^\circ\) and \(\theta = 135^\circ\).