Consider the equation tan x+tan 2x+tan 3x=tan x tan 2x tan 3x for x∈(0,π) at which all the tangents are defined.

Since $\tan3x=\dfrac{\tan x+\tan2x}{1-\tan x\tan2x}$, we have $\tan x+\tan2x=\tan3x\,(1-\tan x\tan2x)$ wherever all three tangents are defined. Substituting, the equation becomes $\tan3x(1-\tan x\tan2x)+\tan3x-\tan x\tan2x\tan3x=0$, i.e. $2\tan3x\,(1-\tan x\tan2x)=0$. The factor $1-\tan x\tan2x=0$ would force $\cos3x=0$, where $\tan3x$ is undefined, so it gives no solution. Hence $\tan3x=0$, i.e. $x=\pi/3$ or $2\pi/3$ in $(0,\pi)$, and at both points $\tan x$ and $\tan2x$ are defined. So there are exactly two solutions (A true). At $x=\pi/6$ and $x=\pi/2$, $\tan3x$ (and at $\pi/2$ also $\tan x$) is undefined, so (B) and (C) are false. The sum of the solutions is $\pi$, not $3\pi/2$, so (D) is false. A fine numerical scan of the original expression shows no other roots.