Answer
The Maclaurin series for $f(x)=\cos(3x)$ is convergent for any value of $x$.
Work Step by Step
By making use of Table 2, we have the Maclaurin series for $f(x)=\cos x$ as follows
$$f(x)=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...$$
Now by the comparison with the function $f(x)=\cos 3x$, we have the Maclaurin series as follows
$$f(x)=\cos 3x=1-\frac{(3x)^2}{2!}+\frac{(3x)^4}{4!}-\frac{(3x)^6}{6!}+...$$
Moreover, again from Table 2, the Maclaurin series for $f(x)=\cos x$ is convergent for any value of $x$. Hence the Maclaurin series for $f(x)=\cos(3x)$ is also convergent for any value of $x$.
