I was wondering if the following equation has a useful property that I can make use of to solve it:
$$ J J^T x = J \cdot g$$ where $g \in \mathbb{R}$ is a scalar and $J \in \mathbb{R}^n$ is a vector.
Is there another way than this:
$$ x = \left( JJ^T\right)^{-1}J\cdot g$$
On
As I understand your question, J is a column vector. If so, then take your equation and multiply both sides by $J^T$. You get $$ J^T J J^T x = J^T J g. $$ Now, unless $J$ is the zero vector, $J^T J$ is a non zero scalar, so your equation simplifies to $$ J^T x = g. $$ This is the equation of a hyperplane in ${\mathbb R}^n$ and hence there are infinitely many solutions (you can, essentially, fill $n-1$ of the entries of $x$ arbitrarily and solve for the other one).
If $J \in \mathbb{R}^n$ then $(J J^T)^{-1}$ will not even exist if $n>1$, so your idea makes no sense.
A better way to look at your equation is just $J^T x=g$, which is a single linear equation in the vector variable $x$. So you can write the general solution to it in the usual way (one variable is dependent, the rest are independent).