Prove that $\sin{\frac{2\pi x}{x^2+x+1}}=\frac{1}{2}$ has no rational roots. 
Show that the following equation has no rational roots.
  $$\sin{\frac{2\pi x}{x^2+x+1}}=\frac{1}{2}$$

This is what I've tried:
$$\left ( \frac{2\pi x}{x^2+x+1}=\frac{\pi}{6}+2k\pi \right)\lor\left (\frac{2\pi x}{x^2+x+1}=\frac{5\pi}{6}+2l\pi \right ), \; \left (k,l \in \mathbb{Z} \right ) $$
First case only:
$$ \frac{2\pi x}{x^2+x+1}=\frac{\pi}{6}+2k\pi \iff \frac{2\pi x}{x^2+x+1}=2\pi \left ( \frac{1}{12}+k \right )$$
$$\left ( 12k+1 \right )x^2+\left (12k-11 \right)x+(12k+1)=0 \tag{$\bigstar$}$$
Now $$\Delta=96k^2-316k+117\geqslant 0 $$
For case $\Delta=0$:
$$k_{1,2}=\frac{316\pm \sqrt{54928}}{2\cdot96} \notin \mathbb{Z}$$
But it means nothing 'cause I still have entire interval $\left ( -\infty,k_{-} \right ) \cup \left ( k_{+}, +\infty\right )$ to check. 
I've rewritten $\left ( \bigstar \right )$:
$$x^2+x\left (\frac{12k-11}{12k+1} \right)+1=0$$
By using Vieta's formulas: $$x_{1}+x_{2}=\frac{11-12k}{12k+1} \qquad \land \qquad x_{1}x_{2}=1$$
Since I have to prove that there are no rational roots, I wanted to give it a shot with contradiction, so I assumed:
$$x_{1}=\frac{p}{q}, \;\; x_{2}=\frac{r}{s}, \;\; \left (p,q \right)=\left (r,s \right)=1$$
Since $\left (p,q \right)=\left (r,s \right)=1$ we obtain from $x_{1}x_{2}=1$ that $p=s$ and $q=r$.  
Now we have: $$x_{1}+x_{2}=\frac{p}{q}+\frac{r}{s}=\frac{p}{q}+\frac{q}{p}$$
$$x_{1}+x_{2}=\frac{p^2+q^2}{pq}=-\frac{12k+1-12}{12k+1}=\frac{12}{12k+1}-1$$ 
Now if I could prove: $$\left (p,q \right)=1 \implies \frac{p^2+q^2}{pq}\in \mathbb{Z}$$
...then I would show that my assumption is wrong i.e. there are no rational roots of the given equation.  
But even if that is true, I'd rather hear some solution that doesn't involve number theory, not even elementary one.

You are almost there. If the equation has rational roots, then $\Delta$ must be a perfect square (why?).
To show that $\Delta$ is not a perfect square just observe that
$$\Delta=96k^2-316k+11 \equiv 3 \pmod{4}$$
