## Precalculus (10th Edition)

$0$
RECALL: $\tan{\theta}=\dfrac{\sin{\theta}}{\cos{\theta}}$ With $\sin{\frac{\pi}{3}} = \frac{\sqrt3}{2}$, $\sin{\frac{\pi}{6}}=\frac{1}{2}$, and $\cos{\frac{\pi}{6}}=\frac{\sqrt3}{2}$, then $2\sin \dfrac {\pi }{3}-3\tan \dfrac {\pi }{6} \\=2\times \left(\dfrac {\sqrt {3}}{2}\right)-3\left(\dfrac {\sin \dfrac {\pi }{6}}{\cos \dfrac {\pi }{6}}\right) \\=\sqrt {3}-3\times \left(\dfrac {\dfrac {1}{2}}{\dfrac {\sqrt {3}}{2}}\right) \\=\sqrt {3}-\sqrt {3} \\=0$