1. Use the identity \(\tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b}\) to express \(\tan(x + 45^\circ)\):

\(\tan(x + 45^\circ) = \frac{\tan x + 1}{1 - \tan x}\)

2. Substitute \(\cot x = \frac{1}{\tan x}\) into the equation:

\(\frac{\tan x + 1}{1 - \tan x} = 2 \cdot \frac{1}{\tan x} + 1\)

3. Clear the fractions by multiplying through by \(\tan x (1 - \tan x)\):

\((\tan x + 1) \tan x = 2(1 - \tan x) + \tan x (1 - \tan x)\)

4. Simplify and rearrange to form a quadratic equation:

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

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

5. Solve the quadratic equation using the quadratic formula \(\tan x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

\(a = 2, \ b = 1, \ c = -1\)

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

\(\tan x = \frac{-1 \pm \sqrt{1 + 8}}{4}\)

\(\tan x = \frac{-1 \pm 3}{4}\)

\(\tan x = \frac{2}{4} = 0.5 \quad \text{or} \quad \tan x = \frac{-4}{4} = -1\)

6. Find \(x\) for \(\tan x = 0.5\):

\(x = \tan^{-1}(0.5) \approx 26.6^\circ\)

7. Find \(x\) for \(\tan x = -1\):

\(x = 135^\circ\)

8. Adjust for the given interval \(0^\circ < x < 180^\circ\):

\(x = 31.7^\circ \quad \text{and} \quad x = 121.7^\circ\)