To solve the equation \(2 \cot 2x + 3 \cot x = 5\), we start by using trigonometric identities.

First, express \(\cot 2x\) using the double angle identity:

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

Substitute \(\cot 2x\) into the equation:

\(2 \left(\frac{1 - \tan^2 x}{2 \tan x}\right) + 3 \cot x = 5\)

Simplify and multiply through by \(\tan x\):

\(1 - \tan^2 x + 3 = 5 \tan x\)

Rearrange to form a quadratic equation:

\(\tan^2 x + 5 \tan x - 4 = 0\)

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

\(y^2 + 5y - 4 = 0\)

Use the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 1\), \(b = 5\), \(c = -4\):

\(y = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1}\) \(y = \frac{-5 \pm \sqrt{25 + 16}}{2}\) \(y = \frac{-5 \pm \sqrt{41}}{2}\)

Calculate the values of \(y\):

\(y_1 = \frac{-5 + \sqrt{41}}{2} \approx 1.702\) \(y_2 = \frac{-5 - \sqrt{41}}{2} \approx -6.702\)

Since \(y = \tan x\), find \(x\) for \(0^\circ < x < 180^\circ\):

\(x_1 = \tan^{-1}(1.702) \approx 35.1^\circ\) \(x_2 = \tan^{-1}(-6.702) \approx 99.9^\circ\)

Thus, the solutions are \(x = 35.1^\circ\) and \(x = 99.9^\circ\).