In (0, π), the number of solutions of the equation ​ tan

(d) 2.

Given equation:

$\tan x+\tan 2 x+\tan 3 x=\tan x \tan 2 x \tan 3 x$

$\Rightarrow \tan x+\tan 2 x=-\tan 3 x+\tan x \tan 2 x \tan 3 x$

$\Rightarrow \tan x+\tan 2 x=-\tan 3 x(1-\tan x \tan 2 x)$

$\Rightarrow \frac{\tan x+\tan 2 x}{1-\tan x \tan 2 x}=-\tan 3 x$

$\Rightarrow \tan (x+2 x)=-\tan 3 x$

$\Rightarrow \tan 3 x=-\tan 3 x$

$\Rightarrow 2 \tan 3 x=0$

$\Rightarrow \tan 3 x=0$

$\Rightarrow 3 x=n \pi$

$\Rightarrow x=\frac{n \pi}{3}$

Now,

$x=\frac{\pi}{3}, n=1$

$x=\frac{2 \pi}{3}, n=2$

$x=\frac{3 \pi}{3}=180^{\circ}$, which is not possible, as it is not in the interval $(0,2 \pi)$

Hence, the number of solutions of the given equation is 2.