Answer
$5-\dfrac{5}{2!}\pi^2x^2+\dfrac{5}{4!}\pi^4x^4-\dfrac{5 }{6!}\pi^6x^6+..$
Work Step by Step
Consider the Maclaurin Series for $\cos x$ as follows:
$ \cos x=\Sigma_{n=0}^\infty \dfrac{(-1)^n x^{2n}}{(2n)!}$
Plug $(\pi x)$ instead of $x$ in the above series.
This implies that
$5 \cos (\pi x)=5 \Sigma_{n=0}^\infty \dfrac{(-1)^n (\pi x)^{2n}}{(2n)!}$
$\implies 5 \cos (\pi x)=5-\dfrac{5}{2!}\pi^2x^2+\dfrac{5}{4!}\pi^4x^4-\dfrac{5 }{6!}\pi^6x^6+..$
