Solve $\tan (\theta) + \tan (2\theta) = \tan (3\theta)$ 
Find the general solution of:
  $$\tan (\theta) + \tan (2\theta) = \tan (3\theta)$$

My Attempt:
$$\tan (\theta) + \tan (2\theta) = \tan (3\theta)$$
$$\dfrac {\sin (\theta)}{\cos (\theta)}+ \dfrac {\sin (2\theta)}{\cos (2\theta)}=\dfrac {\sin (3\theta)}{\cos (3\theta)}$$
$$\dfrac {\sin (\theta+2\theta)}{\cos (\theta) \cos (2\theta)}=\dfrac {\sin (3\theta)}{\cos (3\theta)}$$
 A: \begin{align}
   \tan (\theta) + \tan (2\theta) &= \tan(3\theta) \\
   \tan (\theta) + \tan (2\theta) &= \tan(\theta + 2\theta) \\
   \dfrac{\tan (\theta) + \tan (2\theta)}{1}
      &= \dfrac{\tan(\theta) + \tan(2\theta)}
               {1-\tan (\theta) \tan (2\theta)} \\
\end{align}
So, either $\tan (\theta) + \tan (2\theta)=0$, or 
           $\tan (\theta) = 0$, or $\tan (2\theta)=0$
\begin{align}
   \tan(\theta) + \tan(2\theta) &= 0 \\
   \tan(\theta) + \dfrac{2 \tan(\theta)}{1 - \tan^2(\theta)} &= 0 \\
   3 \tan(\theta) - \tan^3(\theta) &= 0 \\
   \tan(\theta) &\in \{0, \pm \sqrt 3\}
\end{align}
So $\theta \in \left\{
   n\pi, \pm\frac 13\pi + n\pi, \pm\frac 14\pi + n\pi 
   : n \in \mathbb Z
   \right\}$
A: Check if one of $\cos\theta,\cos2\theta,\cos3\theta=0$
Else we have $$\sin3\theta(\cos3\theta-\cos\theta\cos2\theta)=0$$
What if $\sin3\theta=0?$
Else use $$2\cos\theta\cos2\theta=\cos3\theta+\cos\theta$$
A: I'll give a hint to get you started which is that $\text{tan}(a + b) = \dfrac{\text{tan}(a) + \text{tan}(b)}{1-\text{tan}(a)\text{tan}(b)}$.
