Answer
The Maclaurin series for $f(x)=\sin(2x)$ is convergent for any value of $x$.
Work Step by Step
By making use of Table 2, we have the Maclaurin series for $f(x)=\sin x$ as follows
$$f(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...$$
Now by the comparison with the function $f(x)=\sin2x$, we have the Maclaurin series as follows
$$f(x)=\sin2x=2x-\frac{(2x)^3}{3!}+\frac{(2x)^5}{5!}-\frac{(2x)^7}{7!}+...$$
Moreover, again from Table 2, the Maclaurin series for $f(x)=\sin x$ is convergent for any value of $x$. Hence the Maclaurin series for $f(x)=\sin(2x)$ is also convergent for any value of $x$.
