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

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

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

• by general solution, you mean what is $\theta$? – John Glenn Apr 7 '18 at 7:19

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

• $$\tan2\theta=\tan(-\theta)\implies2\theta=n\pi-\theta$$ But what if $\tan2\theta=0$ and $\tan\theta\ne0?$ – lab bhattacharjee Apr 7 '18 at 12:41
• Since $\tan(2\theta)=\dfrac{2 \tan(\theta)}{1-\tan^2(\theta)}$, then that can't happen. – steven gregory Apr 7 '18 at 14:57
• What if $\theta=n\pi+\dfrac\pi2$ – lab bhattacharjee Apr 8 '18 at 1:16
• Then $\tan \theta = \infty$ – steven gregory Apr 8 '18 at 2:11
• and $$\tan2\theta=?$$ – lab bhattacharjee Apr 8 '18 at 5:30

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

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

• @ user2888499, what formula is that? – pi-π Apr 7 '18 at 5:07
• @blue_eyed_... Sorry I had a typo that has been fixed. It's the tangent addition formula. – user2888499 Apr 7 '18 at 5:10
• Whatever the correct formula may be that is clearly wrong. – marty cohen Apr 7 '18 at 5:11
• @martycohen I don't see how this is wrong. I have checked it again for any errors. – user2888499 Apr 7 '18 at 5:13
• @martycohen The only mistake I see is typesetting "tan" as \text{tan} instead of just \tan. The math is correct. – Misha Lavrov Apr 7 '18 at 5:15