We solve the equation \(\cot \theta + \cot(\theta + 45^\circ) = 2\) using the addition formula for tangent.

Let \(t = \tan \theta\). Using the identity

\[ \tan(\theta + 45^\circ) = \frac{\tan \theta + \tan 45^\circ}{1 - \tan \theta \tan 45^\circ} = \frac{t + 1}{1 - t}, \] so \[ \cot(\theta + 45^\circ) = \frac{1}{\tan(\theta + 45^\circ)} = \frac{1 - t}{t + 1}. \]

Also, \(\cot \theta = \frac{1}{t}\). Substituting these into the original equation gives

\[ \frac{1}{t} + \frac{1 - t}{t + 1} = 2. \]

Multiply through by \(t(t+1)\):

\[ (t+1) + t(1 - t) = 2t(t+1). \]

Expanding and simplifying:

\[ t + 1 + t - t^2 = 2t^2 + 2t \quad \Rightarrow \quad 2t - t^2 + 1 = 2t^2 + 2t. \]

Bring all terms to one side:

\[ 0 = 3t^2 - 1 \quad \Rightarrow \quad t^2 = \frac{1}{3}. \]

So

\[ t = \pm \frac{1}{\sqrt{3}}. \]

This gives \(\tan \theta = \frac{1}{\sqrt{3}}\) or \(\tan \theta = -\frac{1}{\sqrt{3}}\).

In degrees:

• \(\tan \theta = \frac{1}{\sqrt{3}}\) gives \(\theta = 30^\circ + 180^\circ k\).
• \(\tan \theta = -\frac{1}{\sqrt{3}}\) gives \(\theta = 150^\circ + 180^\circ k\).

For \(0^\circ < \theta < 180^\circ\), the solutions are \(\boxed{\theta = 30^\circ \text{ and } \theta = 150^\circ}\).