Answer
$\dfrac{7}{2!}x^2+\dfrac{7}{4!}x^4-\dfrac{7}{6!}x^6+..$
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 $(-x)$ instead of $x$ in the above series.
This implies that
$7 \cos (-x)=7 \cos x$
and $7 \Sigma_{n=0}^\infty \dfrac{(-1)^n x^{2n}}{(2n)!}=\dfrac{7}{2!}x^2+\dfrac{7}{4!}x^4-\dfrac{7}{6!}x^6+..$
