Finding solution of a Trigonometric equationtan⁡A+tan⁡2A+tan⁡3A=0I tried converting these all in sin and cos and...

${\displaystyle \tan A+\tan 2A=\frac{\sin 3A}{\cos A\cos 2A}}$
$\tan A+\tan 2A+\tan 3A=0\Rightarrow \frac{\sin 3A}{\cos A\cos 2A}+\frac{\sin 3A}{\cos 3A}=0$
$\iff \sin 3A(\cos 3A+\cos A\cos 2A)=0$
If ${\displaystyle \sin 3A=0,3A=n\pi }$ where n is any integer
Otherwise,
${\displaystyle \cos 3A+\cos A\cos 2A=0\iff 4\cos 3A-3\cos A+\cos A(2{\cos }^{2}A-1)=0}$
${\displaystyle \iff \cos A(6{\cos }^{2}A-4)=0}$
If ${\displaystyle \cos A=0,A=(2m+1)\frac{\pi }{2}}$
Otherwise, ${\displaystyle 6{\cos }^{2}A-4=0\iff {\cos }^{2}A=\frac{23}{,}\cos 2A=2{\cos }^{2}A-1=\frac{13}{}}$

Hint
The following identities for the tangent of multiple angles should be useful
$\tan 2A=\frac{2\tan A}{1-2{\tan }^{2}A}$
$\tan 3A=\frac{3\tan A-{\tan }^{3}A}{1-3{\tan }^{2}A}$
I am sure that you can take from here.