Another beginners calculus problem

Find an $\displaystyle M>0$ so that every $\displaystyle x>M$ fulfills $\displaystyle \frac{6x^2-7\sin x}{3x+1} > 1000$

I've been brainstorming ideas, but it's a bit frustrating since I can't simplify the equation like I usually do.

Here are my thoughts and questions:

1) I'm thinking along the lines of proving something like $\displaystyle \frac{6x^2-7\sin x}{3x+1} > \frac{9x^2-1}{3x+1} > 1000$ because then I can simplify to $\displaystyle 3x-1>1000$. Is there any way I can make that happen?

2) Can $\displaystyle 6x^2-7\sin x$ ever be $\displaystyle < 0$. Because if not, we know that $\displaystyle 3x+1$ must also be$\displaystyle > 0$ since the function must be $\displaystyle > 1000$.

3) Or can I just say M = 1000000000000, and since sinx is always between 1 and -1, it doesn't effect us, and 6x^2 is always bigger than 3x+1. so a x>M is definitely bigger than 1000.

Thanks!