How to approximate $\pi$ using the Maclaurin series for $\sin(x)$

I'd rather work with $\dfrac\pi6$ for a faster converging series.

Then you can indeed solve for the roots of

$$\sin\frac\pi6=\frac12=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$$

by Newton and progressively increase the number of terms and the accuracy with which you compute the roots.

For instance, the given quintic equation yields $\pi\approx3.1415778790776$, and the next approximation is $3.1415927099226$.

I don't think that this is a very efficient method as it makes you evaluate the sine for many values using a (truncated) series during the resolution of the equations, instead of a direct evaluation, for instance by the Machin formula.

It is easy to find better approximations to $\pi$ by iteration. Let $a_{n+1}=a_n+\sin(a_n)$. If you start with $a_1$ close enough to $\pi$ the sequence converges to $\pi$. For example, $a_1=3$ will work. You can replace $\sin(x)$ with a truncated Taylor series $f(x)$ and the iteration will converge to the root of $f(x)$ closest to $\pi$.