How to solve $7^x -5x^3 \equiv 0 \quad \pmod{11}?$ I have to study the solvability of the equation $$ 7^x -5x^3 \equiv 0 \quad \pmod{33} $$ and determine its integer solutions $ x $ with $ 0 \le x \le 110 $.
I started dividing this equation into two equations $$\cases {7^x -5x^3 \equiv 0 \quad \pmod{3} \\ 7^x -5x^3 \equiv 0 \quad \pmod{11}}.$$ For the first one I tried to substitute the values $0,1,2 $ and found that $2$ is the only possibile solution. 
Then I tried to solve the second one with the method of indices: 
$$ x \cdot\mbox{ind}_{11}(7)-\mbox{ind}_{11}(5)-3\cdot\mbox{ind}_{11}(x) \equiv 0 \pmod{\phi(11)}.$$ I noticed that $2$ is a primitive root $\pmod{11}$ and computed its powers which lead to $$2^4 \equiv 5 \pmod{11} \\2^7 \equiv 7 \pmod{11}.$$
The equation became $$ x \cdot 7 - 4- 3 \cdot \mbox{ind}_{11}(x) \equiv 0 \pmod{\phi(11)} \quad \longrightarrow \quad 7 x - 3 \cdot \mbox{ind}_{11}(x) \equiv 4 \quad \pmod{\phi(11)}.$$
Then I stopped because I had no clues on how to continue.
Have you any ideas?
Thank you in advance.

Instead of using the primitive root $2$, note that $7$ is a primitive root mod 11, because $7^2\equiv 5$ and $7^5\equiv -1$.  So using 7-indices is a sensible move.
We list the values:
$$
\begin{array}{c|c|c}
\hline
x\pmod{11} & 5x^3\equiv 7^x\pmod{11} & x\pmod{10}
\\\hline
1 & 5&2\\
2 & 7&1\\
3 & 3&4\\
4 & 1&0\\
5 & 9&8\\
6 & 2&3\\
7 &10&5\\
8 & 8&9\\
9 & 4&6\\
10& 6&7\\\hline
\end{array}
$$
So combining the outer columns, you know ten solutions $x\pmod{110}$ and you know $x\equiv 2\pmod{3}$.  Hence you can write down the ten solutions mod 330, and determine the solutions between 0 and 110.
