Given that$\displaystyle \tan(x) = \frac{2}{3}\mbox{ and } \pi<x<\frac{3\pi}{2} \mbox{, find }\sin(x)\\$

I know that $\displaystyle \pi<x<\frac{3\pi}{2}$ puts $\displaystyle x$ in Quadrant III in the unit circle, but I don't see how that helps me, other than the fact that $\displaystyle \sin(x)$ will be negative. I solved for $\displaystyle \sin(x)$, but I'm pretty sure it isn't the right answer.

$\displaystyle \tan(x) = \frac{2}{3}\\\\\frac{\sin(x)}{cos(x)} = \frac{2}{3}\\\\\sin(x) = \frac{2cos(x)}{3}\\$

also, it gave a,b,c,d,e choices.

a) $\displaystyle \frac{\sqrt{10}}{10}$

b) $\displaystyle \frac{-\sqrt{13}}{2}$

c) $\displaystyle \frac{-\sqrt{10}}{10}$

d) $\displaystyle \frac{-2\sqrt{13}}{13}$

e) $\displaystyle \frac{2\sqrt{13}}{13}$