If it helps, you can replace [tex]\tan x[/tex] with [tex]y[/tex]. How would you go about solving [tex]y^2-y=0[/tex]?
We can write
[tex]\tan x(\tan x-1)=0[/tex]
from which we have two possibilities, either [tex]\tan x=0[/tex] or [tex]\tan x-1=0[/tex].
Both equations have infinitely many solutions because [tex]\tan x=\tan(x+n\pi)[/tex] for any integer [tex]n[/tex]. But we're viewing [tex]x[/tex] as a positive angle, which means [tex]0<x<2\pi[/tex]. Moreover, we can assume [tex]x[/tex] is an acute angle, so that [tex]0<x<\dfrac\pi2[/tex].
Now, [tex]\tan x=0[/tex] for [tex]x=n\pi[/tex], which means there are no solutions to this equation on this interval.
On the other hand, [tex]\tan x=1[/tex] for [tex]x=\dfrac\pi4[/tex].
