Continue applying the tangent addition formula.

From the previous step, we have:

$$\frac{\tan \theta + \tan 2\theta}{1 - \tan \theta \tan 2\theta} = 1$$

The left side of this equation is the expansion of $\tan(\theta + 2\theta)$ using the tangent addition formula $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$.
So, we can write:

$$\tan(\theta + 2\theta) = 1$$ $$\tan(3\theta) = 1$$

The general solution for an equation of the form $\tan x = k$ is $x = n\pi + \arctan(k)$, where $n$ is an integer.
In this case, $k=1$, so $\arctan(1) = \frac{\pi}{4}$.
Therefore, for $\tan(3\theta) = 1$:

$$3\theta = n\pi + \frac{\pi}{4}$$

Divide by 3 to solve for $\theta$:

$$\theta = \frac{n\pi}{3} + \frac{\pi}{12}$$

where $n \in \mathbb{Z}$ (the set of all integers).

The final answer is $\boxed{\theta = \frac{n\pi}{3} + \frac{\pi}{12}, \text{ where } n \in \mathbb{Z}}$.