Because $11$ is a prime, any solution must be co-prime to $11$. Then by Fermat's little theorem, we have $x^{10}\equiv 1\pmod{11}$. Then it follows that $x^2\equiv 3\pmod{11}$. We test $x=0$, $\pm 1$, $\pm 2$, $\pm 3$, $\pm 4$, $\pm 5$ and find $x\equiv \pm 5$ are the solutions.