<s>[INST] 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.
 [/INST] $$\tan(a) + \tan(2a) + \tan(3a) = 0 $$
$$\tan(2a) = 2 \cdot \frac{\tan(a)}{ (1 - \tan(a)^2) }$$
$$\tan(3a) =\tan(2a + a)$$ 
$$= \frac{\tan(a) + \tan(2a }{ 1 - \tan(a) \cdot \tan(2a)}$$ 
$$= \frac{(\tan(a) + \frac{2 \cdot \tan(a }{1 - \tan(a)^2)} }{ \frac{(1 - tan(a) \cdot 2 \cdot \tan(a) }{ 1 - \tan(a)^2}}$$ 
$$\frac{\tan(a) - \tan(a)^3 + 2\tan(a) }{(1 - \tan(a)^2 - 2\tan(a)^2}$$ 
$$ = \frac{3\tan(a) - \tan(a)^3}{ 1 - 3\tan(a)^2}  $$
$$=\tan(a) \cdot \frac{3 - \tan(a)^2 }{ 1 - 3\tan(a)^2} $$
Given
$$\tan(a) + \tan(2a) + \tan(3a) = 0 $$
$$\tan(a) + \frac{2\tan(a) }{ (1 - \tan(a)^2)} + \frac{\tan(a) \cdot (3 - \tan(a)^2) }{ (1 - 3\tan(a)^2)) } = 0$$
$$\tan(a) = 0 $$
$$a = \pi \cdot k $$
$$1 +\frac{2}{(1 -\tan(a)^2)}+\frac{(3 -\tan(a)^2) }{(1 - 3\tan(a)^2)} = 0 $$
$$\frac{((1 - \tan(a)^2) \cdot (1 - 3\tan(a)^2) + 2\cdot (1-3\tan(a)^2) + (3 -\tan(a)^2) \cdot (1 -\tan(a)^2)) }{ ((1 -\tan(a)^2)\cdot (1-3\tan(a)^2)) }= 0$$ 
$$(1 - \tan(a)^2) \cdot (1 - 3\tan(a)^2) + 2 \cdot (1 - 3\tan(a)^2) + (3 - \tan(a)^2) \cdot (1 - \tan(a)^2) = 0 $$
$$1 - 4\tan(a)^2 + 3\tan(a)^4 + 2 - 6\tan(a)^2 + 3 - 4\tan(a)^2 + \tan(a)^4=0 $$ 
$$6 - 14\tan(a)^2 + 4\tan(a)^4 = 0 $$
$$2\tan(a)^4 - 7\tan(a)^2 + 3 = 0 $$
$$\tan(a)^2 = \frac{(7 \pm \sqrt{(49 - 24}}{ 4 }$$
$$\tan(a)^2 = \frac{(7 \pm 5)}{4} $$
$$\tan(a)^2 = \frac{12}4 , \frac24 $$
$$\tan(a)^2 = 3 , \frac1 2 $$
$$\tan(a) = \pm \sqrt(3) , \pm \sqrt\frac{(2)}{ 2} $$
$$\tan(a) = 0, -\sqrt(3), \sqrt(3), -\sqrt\frac{(2)}{ 2}, \sqrt\frac{(2)}{ 2}$$
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