Answer
See explanations.
Work Step by Step
Step 1. Using the Addition Formula, we have
$sin(x+\frac{\pi}{6})=sin(x)cos(\frac{\pi}{6})+cos(x)sin(\frac{\pi}{6})=\frac{\sqrt 3}{2}sin(x)+\frac{1}{2}cos(x)$
and
$cos(x+\frac{\pi}{3})=cos(x)cos(\frac{\pi}{3})-sin(x)sin(\frac{\pi}{3})=\frac{1}{2}cos(x)-\frac{\sqrt 3}{2}sin(x)$
Step 2. $LHS=\frac{\sqrt 3}{2}sin(x)+\frac{1}{2}cos(x)+\frac{1}{2}cos(x)-\frac{\sqrt 3}{2}sin(x)=cos(x)=RHS$
Step 3. Since $LHS=RHS$, we verified the identity.
