3.328.

$$
\frac{\cos \left(\alpha+\frac{3}{2} \pi\right)+2 \cos \left(\frac{11}{6} \pi-\alpha\right)}{2 \sin \left(\frac{\pi}{3}+\alpha\right)+\sqrt{3} \sin \left(\frac{3}{2} \pi-\alpha\right)}
$$
## Solution.

$$
\begin{aligned}
& \frac{\cos \left(\alpha+\frac{3}{2} \pi\right)+2 \cos \left(\frac{11}{6} \pi-\alpha\right)}{2 \sin \left(\frac{\pi}{3}+\alpha\right)+\sqrt{3} \sin \left(\frac{3}{2} \pi-\alpha\right)}=\frac{\cos \left(\frac{3}{2} \pi+\alpha\right)+2 \cos \left(2 \pi-\left(\frac{\pi}{6}+\alpha\right)\right)}{2 \sin \left(\frac{\pi}{3}+\alpha\right)+\sqrt{3} \sin \left(\frac{3}{2} \pi-\alpha\right)}= \\
& =\frac{\sin \alpha+2 \cos \left(\frac{\pi}{6}+\alpha\right)}{2 \sin \left(\frac{\pi}{3}+\alpha\right)-\sqrt{3} \cos \alpha}=[\cos (x+y)=\cos x \cos y-\sin x \sin y \\
& \sin (x+y)=\sin x \cos y+\cos x \sin y]= \\
& =\frac{\sin \alpha+2 \cos \frac{\pi}{6} \cos \alpha-2 \sin \frac{\pi}{6} \sin \alpha}{2 \sin \frac{\pi}{3} \cos \alpha+2 \cos \frac{\pi}{3} \sin \alpha-\sqrt{3} \cos \alpha}=
\end{aligned}
$$

$=\frac{\sin \alpha+\sqrt{3} \cos \alpha-\sin \alpha}{\sqrt{3} \cos \alpha+\sin \alpha-\sqrt{3} \cos \alpha}=\frac{\sqrt{3} \cos \alpha}{\sin \alpha}=\sqrt{3} \operatorname{ctg} \alpha$.

Answer: $\sqrt{3}$ ctg $\alpha$.