How to find the roots of sin(x) using series theory

If we define $$\sin(x) = \sum_{n=0}^\infty \frac{(-1)^n \ x^{2n+1}}{(2n+1)!}$$How to find the roots of $\sin(x)$, i.e.$$\pi =4 \sum_{n=0}^\infty \frac{(-1)^n}{2n+1}$$satisfies $\sin (\pi)=0$