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.
 A: 
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.
A: Maybe not the prettiest way to solve this, but dealing with sine and cosine is usually easier, so if $\tan(ax)=\frac{\sin(2ax)}{cos(2ax)+1}$, then we have a sum:
$$
\sum_{a=1}^{a=3}\frac{\sin(2ax)}{cos(2ax)+1}
$$
which after algebra becomes
$$
\frac{\sin(3x) \sec(x) \sec(2x)\ (3\cos(x)-1)}{2\cos(2x)-1}
$$
which may be solved by setting
$$
\sin(3x) \sec(x) \sec(2x)\ (3\cos(x)-1) = 0.
$$
A: Hint:
$$\sin(A+2A)\cos3A+\sin3A\cos A\cos2A=0$$
What if $\sin3A=0?$
Else $$0=2\cos3A+2\cos A\cos2A=2\cos3A+\cos(2A-A)+\cos(2A+A)=?$$
A: $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
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 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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$\ds{\tan\pars{A} + \tan\pars{2A} + \tan\pars{3A} = 0\,,\qquad
A \in \bracks{0,2\pi}}$.

Lets $\ds{\tan\pars{A} \equiv x}$ such that
$\ds{\tan\pars{2A} = {2\tan\pars{A} \over 1 - \tan^{2}\pars{A}} =
{2x \over 1 - x^{2}}}$.

\begin{align}
0 & = \tan\pars{A} + \tan\pars{2A} + \tan\pars{3A} =
x + {2x \over 1 - x^{2}} + {\tan\pars{2A} +
\tan\pars{A} \over 1 - \tan\pars{2A} \tan\pars{A}}
\\[5mm] & =
x + {2x \over 1 - x^{2}} + {2x/\pars{1 - x^{2}} + x \over
1 - \bracks{2x/\pars{1 - x^{2}}}x} =
{2x\pars{2x^{4} - 7x^{2} + 3} \over 3x^{4} - 4x^{2} + 1}
\end{align}

Then,
\begin{equation}
x = \tan\pars{A}\ \in\ \Omega \equiv\braces{\pm\root{3},\pm{\root{2} \over 2},0}
\label{1}\tag{1}
\end{equation}


From \eqref{1}, you can deduce the values of $\ds{A \in \bracks{0,2\pi}}$. Note that
  $\ds{\pars{3x^{4} - 4x^{2} + 1}_{\ x\ \in\ \Omega} \not= 0}$.

A: $$\tan(a) + \tan(2a) + \tan(3a) = 0 $$
$$\tan(2a) = 2 \cdot \frac{\tan(a)}{ (1 - \tan(a)^2) }$$
$$\tan(3a) =\tan(2a + a)$$ 
$$= \frac{\tan(a) + \tan(2a }{ 1 - \tan(a) \cdot \tan(2a)}$$ 
$$= \frac{(\tan(a) + \frac{2 \cdot \tan(a }{1 - \tan(a)^2)} }{ \frac{(1 - tan(a) \cdot 2 \cdot \tan(a) }{ 1 - \tan(a)^2}}$$ 
$$\frac{\tan(a) - \tan(a)^3 + 2\tan(a) }{(1 - \tan(a)^2 - 2\tan(a)^2}$$ 
$$ = \frac{3\tan(a) - \tan(a)^3}{ 1 - 3\tan(a)^2}  $$
$$=\tan(a) \cdot \frac{3 - \tan(a)^2 }{ 1 - 3\tan(a)^2} $$
Given
$$\tan(a) + \tan(2a) + \tan(3a) = 0 $$
$$\tan(a) + \frac{2\tan(a) }{ (1 - \tan(a)^2)} + \frac{\tan(a) \cdot (3 - \tan(a)^2) }{ (1 - 3\tan(a)^2)) } = 0$$
$$\tan(a) = 0 $$
$$a = \pi \cdot k $$
$$1 +\frac{2}{(1 -\tan(a)^2)}+\frac{(3 -\tan(a)^2) }{(1 - 3\tan(a)^2)} = 0 $$
$$\frac{((1 - \tan(a)^2) \cdot (1 - 3\tan(a)^2) + 2\cdot (1-3\tan(a)^2) + (3 -\tan(a)^2) \cdot (1 -\tan(a)^2)) }{ ((1 -\tan(a)^2)\cdot (1-3\tan(a)^2)) }= 0$$ 
$$(1 - \tan(a)^2) \cdot (1 - 3\tan(a)^2) + 2 \cdot (1 - 3\tan(a)^2) + (3 - \tan(a)^2) \cdot (1 - \tan(a)^2) = 0 $$
$$1 - 4\tan(a)^2 + 3\tan(a)^4 + 2 - 6\tan(a)^2 + 3 - 4\tan(a)^2 + \tan(a)^4=0 $$ 
$$6 - 14\tan(a)^2 + 4\tan(a)^4 = 0 $$
$$2\tan(a)^4 - 7\tan(a)^2 + 3 = 0 $$
$$\tan(a)^2 = \frac{(7 \pm \sqrt{(49 - 24}}{ 4 }$$
$$\tan(a)^2 = \frac{(7 \pm 5)}{4} $$
$$\tan(a)^2 = \frac{12}4 , \frac24 $$
$$\tan(a)^2 = 3 , \frac1 2 $$
$$\tan(a) = \pm \sqrt(3) , \pm \sqrt\frac{(2)}{ 2} $$
$$\tan(a) = 0, -\sqrt(3), \sqrt(3), -\sqrt\frac{(2)}{ 2}, \sqrt\frac{(2)}{ 2}$$
