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

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, $\forall k\in\mathbb{Z}$
$$x=k\pi\lor x=\pm \frac{\pi}{3}+k\pi \lor x=\pm\arctan\left(\sqrt{\frac{1}{2}}\right)+k\pi$$
