Find general solution for $\tan{x} +\tan{2x} +\tan{3x}= 0$ My approach :-
$\tan{(x+2x)}= \tan{3x}$
$\tan{x}+\tan{2x} = \tan{3x}(1-\tan{x}.\tan{2x})$
Putting this in my equation ,
$\tan{3x}.(2-\tan{x}tan{2x}) =0$
Now either
$\tan{3x}= 0$    or  $\tan{x}\tan{2x}=2$
Hence we get either
$x=\frac{n\pi}{3}$    or  $x= n\pi ± \arctan{\frac{1}{√2}}.$
However , my textbook says that the solution is 
$x=n\pi$    or    $x= n\pi \pm \frac{\pi}{3}$
Why is there a disparity in both the solutions? If they are equivalent , how?
I apologize in advance as I'm using latex for the first time. Thank You
 A: $$\tan x+\tan2x=\dfrac{\sin(x+2x)}{\cos x\cos2x}$$
$$\implies\sin3x(\cos3x+\cos x\cos2x)=0$$ with $\cos x\cos2x\cos3x\ne0$
What if $\sin3x=0?$
Else $$-\cos3x=\cos x\cos2x$$
Now $\cos3x=\cos x(4\cos^2x-3)=\cos x(2(1+\cos2x)-3)$
As $\cos x\ne0,$
$$\cos2x=-(2\cos2x-1)$$
Can you take it from here to reach at your second solution
A: I couldn't follow your approach, so I'll show my own. You'll want to double-check all the calculations.
Define $t:=\tan x$ so $\tan 2x=\frac{2t}{1-t^2},\,\tan 3x=\frac{t(3-t^2)}{1-3t^2}$ and $\tan x+\tan 2x+\tan 3x=\frac{2t(3-t^2)(1-2t^2)}{\left(1-t^{2}\right)\left(1-3t^{2}\right)}$. This is $0$ iff $t\in\left\{0,\,\pm\frac{1}{\sqrt{2}},\,\pm\sqrt{3}\right\}$.
A: Remember
$$\tan (2 x)=\frac{2 \tan (x)}{1-\tan ^2(x)}$$
and
$$\tan(3x)=\frac{\tan (x) \left(\tan ^2(x)-3\right)}{3 \tan ^2(x)-1}$$
So the given equation can be written as
$$\frac{2 \tan (x)}{1-\tan ^2(x)}+\frac{\left(\tan ^2(x)-3\right) \tan (x)}{3 \tan ^2(x)-1}+\tan (x)=0$$
Adding the fractions
$$\frac{2 \tan (x) \left(\tan ^2(x)-3\right) \left(2 \tan ^2(x)-1\right)}{\left(\tan ^2(x)-1\right) \left(3 \tan ^2(x)-1\right)}=0$$
which is verified when
$$\tan(x)=0\lor \tan ^2(x)-3=0\lor 2 \tan ^2(x)-1=0$$
that is, $\forall k\in\mathbb{Z}$
$$x=k\pi\lor x=\pm \frac{\pi}{3}+k\pi \lor x=\pm\arctan\left(\sqrt{\frac{1}{2}}\right)+k\pi$$
A: Let $t=\tan x$ and $T=\tan2x$. Then 
$$\tan x+\tan2x+\tan3x=t+T+{T+t\over1-Tt}=(T+t)\left({2-Tt\over1-Tt} \right)$$
So $\tan x+\tan2x+\tan3x=0$ if and only if $T+t=0$ or $2-Tt=0$. Using $T=2t/(1-t^2)$, these become (after a little algebra)
$$t(3-t^2)=0\quad\text{or}\quad1-2t^2=0$$
from which we have $\tan x=0$, $\pm\sqrt3$, or $\pm1/\sqrt2$.
