Diophantine approximation and asymptotic for $\dfrac{1}{\sin(n\pi\sqrt{3})}$

I have this exercise and i proved the lemma but i couldn't use it to prove the asymptotic formula i couldn't plug the sin function into the inequality because it change variations maybe some choice of p and q is the key but not sure what to do.

The problem: Show that as $n$ goes to $+\infty$,$$ \dfrac{1}{\sin(n\pi\sqrt{3})} = O(n) $$

Start by proving the lemma : $\exists c > 0$,

$$\left\vert \dfrac{p}{q} - \sqrt{3} \right\vert \geq\dfrac{c}{q^{2} }$$

for all rational numbers $\dfrac{p}{q}$