#### Answer

$\dfrac{1}{2}$

#### 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)!}=1-\dfrac{1}{2!}x^2+\dfrac{1}{4!}x^4-...$
As we are given that $x=\dfrac{\pi}{3}$
Then, we have
$\cos (x)=\cos (\dfrac{\pi}{3})=\dfrac{1}{2}$