To solve \(\cot 2\theta = 2 \tan \theta\), we start by using trigonometric identities:

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

Equating this to \(2 \tan \theta\):

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

Cross-multiply to clear the fraction:

\(1 - \tan^2 \theta = 4 \tan^2 \theta\)

Rearrange the equation:

\(1 = 5 \tan^2 \theta\)

\(\tan^2 \theta = \frac{1}{5}\)

Take the square root:

\(\tan \theta = \pm \frac{1}{\sqrt{5}}\)

Calculate \(\theta\) using \(\tan^{-1}\):

\(\theta = \tan^{-1} \left( \frac{1}{\sqrt{5}} \right) \approx 24.1^\circ\)

For the second solution in the range \(0^\circ < \theta < 180^\circ\):

\(\theta = 180^\circ - 24.1^\circ = 155.9^\circ\)

Thus, the solutions are \(\theta = 24.1^\circ\) and \(\theta = 155.9^\circ\).