To solve \(\tan 2x = 5 \cot x\), we start by using the identity \(\cot x = \frac{1}{\tan x}\).

Thus, the equation becomes:

\(\tan 2x = \frac{5}{\tan x}\)

Using the double angle formula for tangent, \(\tan 2x = \frac{2 \tan x}{1 - \tan^2 x}\), we substitute:

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

Cross-multiplying gives:

\(2 \tan^2 x = 5(1 - \tan^2 x)\)

Expanding and rearranging terms, we get:

\(2 \tan^2 x = 5 - 5 \tan^2 x\)

\(7 \tan^2 x = 5\)

\(\tan^2 x = \frac{5}{7}\)

Taking the square root, we find:

\(\tan x = \pm \sqrt{\frac{5}{7}}\)

Now, solve for \(x\) in the interval \(0^\circ < x < 180^\circ\).

For \(\tan x = \sqrt{\frac{5}{7}}\), we find:

\(x = \tan^{-1}\left(\sqrt{\frac{5}{7}}\right) \approx 40.2^\circ\)

For \(\tan x = -\sqrt{\frac{5}{7}}\), we find:

\(x = 180^\circ - \tan^{-1}\left(\sqrt{\frac{5}{7}}\right) \approx 139.8^\circ\)

Thus, the solutions are \(x = 40.2^\circ\) and \(x = 139.8^\circ\).