Answer
The identity is verified.
$cos(\frac{\pi}{3}+x)+cos(\frac{\pi}{3}-x)=cos~x$
Work Step by Step
$cos(\frac{\pi}{3}+x)+cos(\frac{\pi}{3}-x)=2~cos(\frac{\frac{\pi}{3}+x+\frac{\pi}{3}-x}{2})~cos(\frac{\frac{\pi}{3}+x-(\frac{\pi}{3}-x)}{2})=2~cos\frac{\pi}{3}~cos~x=2(\frac{1}{2})cos~x=cos~x$
