Any solution $x$ must be co-prime to $11$. By Fermat's little theorem, we have $x^{10}\equiv 1\pmod{11}$. By Euclid's algorithm, we have $$3\times 7 - 2\times 10 = 1$$

It follows that $$3^3 \equiv \left(x^7\right)^3\equiv \left(x^{10}\right)^2\cdot x\equiv x\pmod{11}$$

Hence we find one solution $x\equiv 3^3\equiv \boxed{5}\pmod{11}$.