The set of values of x satisfying the equation

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

i.e $\tan (3 x-2 x)=1 \quad$ (using identity :- $\left.\tan (x-y)=\frac{\tan x-\tan y}{1+\tan x \tan y}\right)$

i.e $\tan x=1$

i.e $x=\tan ^{-1} \frac{\pi}{4}$

i.e $x=n \pi+\frac{\pi}{4} \quad ; n \in Z$