Answer
$\cos(x+\frac{\pi}{3})+\sin(x-\frac{\pi}{6})=0$
Work Step by Step
Start with the left side:
$\cos(x+\frac{\pi}{3})+\sin(x-\frac{\pi}{6})$
Expand using the addition formula for cosine and the subtraction formula for sine:
$=(\cos x\cos \frac{\pi}{3}-\sin x\sin \frac{\pi}{3})+(\sin x\cos\frac{\pi}{6}-\cos x\sin \frac{\pi}{6})$
Evaluate $\cos \frac{\pi}{3}$, $\sin \frac{\pi}{3}$, $\cos\frac{\pi}{6}$, and $\sin \frac{\pi}{6}$:
$=(\cos x*\frac{1}{2}-\sin x*\frac{\sqrt{3}}{2})+(\sin x*\frac{\sqrt{3}}{2}-\cos x*\frac{1}{2})$
Simplify:
$=\frac{1}{2}\cos x-\frac{\sqrt{3}}{2}\sin x+\frac{\sqrt{3}}{2}\sin x-\frac{1}{2}\cos x$
$=0$
