If I have the equation:
$x=-$tan$x$
I can send $x \rightarrow -x$ and the equation doesn't change. If I define $x>0$ one of the possible ranges $x$ can take without solving the equation is
$\pi/2<x<\pi$
because of invariance if $x<0$, one of the possible ranges $x$ can take is:
$-\pi<x<-\pi/2$
However if I were to have an invariant equation in which there was no solution for $x>0$, would that necessarily mean there is no solution to the equation for $x<0$?