Find all $0^\circ\leq A\leq360^\circ$ with $\tan A + \tan 2A + \tan 3A = 0$ solve:
$\tan A+ \tan 2A+ \tan 3A=0$
My Attempt:
$$\tan A+\tan 2A+\tan 3A=0$$
$$\tan A+\frac {2\tan A}{1-\tan^2A}+\frac {3\tan A-\tan^3A}{1-3\tan^2A}=0$$
$$\frac {\tan A-\tan^3A+2\tan A}{1-\tan^2A}+\frac {3\tan A-\tan^3A}{1-3\tan^2A}=0$$
What should I do further. Please suggest.


What should I do further. Please suggest.

Multiply the both sides by $(1-\tan^2A)(1-3\tan^2A)$ and factorize the left-hand side.

The following way might be easier.
Since
$$-\tan A-\tan(2A)=\tan(3A)=\tan(A+2A)=\frac{\tan A+\tan(2A)}{1-\tan A\tan(2A)}$$
we get
$$(\tan A+\tan(2A))\left(1+\frac{1}{1-\tan A\tan(2A)}\right)=0$$
$$\tan A+\tan(2A)=0\quad\text{or}\quad \tan A\tan(2A)=2$$
$$t+\frac{2t}{1-t^2}=0\quad\text{or}\quad \frac{2t^2}{1-t^2}=2$$
where $t=\tan A$, which should be easy to deal with.
