Find all positive integers $x$, that satisfy $29x^{33} \equiv 27\pmod {11} $ 
Find all positive integers $x$, that satisfy $29x^{33} \equiv 27 \pmod {11}$.

I approached this the following way:
Since from $29x^{33} \equiv 27 \pmod {11}$  we get that $7x^{33} \equiv 5 \pmod {11}$ and since $\gcd(7,5)=1$ we would get that $\phi(11)=10$ which would imply that $7x^{10} \equiv 5 \pmod {11}$.
How should I continue from here, it doesn't seem to be quite clear.

Another path:  Subtract $11$ from $29$ to get
$$18x^3\equiv 27 \pmod{11}.$$
Divide by $9$
$$2x^3 \equiv 3 \equiv 14 \pmod{11}.$$
Divide by $2$
$$x^3 \equiv 7 \pmod{11}.$$
Cube both sides
$$x^9 \equiv x^{-1} \equiv 49\cdot 7 \equiv 2 \pmod{11}.$$
Solve $2x\equiv 1 \pmod{11}$ to get $x=6$.
