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.

$$29\equiv7,27\equiv5\pmod{11}$$
and $$\phi(11)=10$$
$$\implies7x^3\equiv5\pmod{11}$$
Now as $7\cdot8\equiv1\pmod{11},$
$$ x^3\equiv5\cdot8\equiv(-4)\pmod{11}$$
Finally as $1=10-3\cdot3,$
$$x=x^{10}(x^3)^{-3}\equiv1\cdot(-4)^{-3}\equiv-2^{-6}\equiv-2^4\equiv6$$
