Find x in the congruence using Fermat's Little Theorem

It's supposed to be 2^(11), but don't know how to do it(Speechless)

$\displaystyle 2^11 \equiv xmod3$ such that $\displaystyle 0<=x<=2$

Find x using Fermat's Little Theorem.

Here is what I have done:

Since 3 is prime, it follows that $\displaystyle (2^11)^2 \equiv 1mod3$ by FLT and since $\displaystyle (2^11,3)=1$.

So, multiply both sides by $\displaystyle 2^11$ we get

$\displaystyle (2^11)^2 \equiv 2^11 * xmod3$

$\displaystyle 1 \equiv 2^11 * xmod3$

It is here I'm not sure what to do since not sure how to get something $\displaystyle 0<=x<=2$, or how to isolate the x really.

Thanks