Continue applying the tangent addition formula.

Step 1: 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: $$\tan(\theta + 2\theta) = 1$$ $$\tan(3\theta) = 1$$ Step 2: Find the general solution for $\theta$. 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 $\boxed{\theta = \frac{n\pi}{3} + \frac{\pi}{12}, \text{ where } n \in \mathbb{Z}}$. Last free one today — make it count tomorrow, or type /upgrade for unlimited.