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

Question

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

My Attempt: $$\tan (\theta) + \tan (2\theta) = \tan (3\theta)$$ $$\dfrac {\sin (\theta)}{\cos (\theta)}+ \dfrac {\sin (2\theta)}{\cos (2\theta)}=\dfrac {\sin (3\theta)}{\cos (3\theta)}$$ $$\dfrac {\sin (\theta+2\theta)}{\cos (\theta) \cos (2\theta)}=\dfrac {\sin (3\theta)}{\cos (3\theta)}$$

Answer 1

\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 $\theta \in \left\{ n\pi, \pm\frac 13\pi + n\pi, \pm\frac 14\pi + n\pi : n \in \mathbb Z \right\}$

Answer 2

I'll give a hint to get you started which is that $\text{tan}(a + b) = \dfrac{\text{tan}(a) + \text{tan}(b)}{1-\text{tan}(a)\text{tan}(b)}$.

Answer 3

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 $$2\cos\theta\cos2\theta=\cos3\theta+\cos\theta$$