3.245.

$$
\frac{\sin (2 \alpha-3 \pi)+2 \cos \left(\frac{7}{6} \pi+2 \alpha\right)}{2 \cos \left(\frac{\pi}{6}-2 \alpha\right)+\sqrt{3} \cos (2 \alpha-3 \pi)}
$$
## Solution.

$$
\begin{aligned}
& \frac{\sin (2 \alpha-3 \pi)+2 \cos \left(\frac{7}{6} \pi+2 \alpha\right)}{2 \cos \left(\frac{\pi}{6}-2 \alpha\right)+\sqrt{3} \cos (2 \alpha-3 \pi)}= \\
& =\frac{-\sin (3 \pi-2 \alpha)+2 \cos \left(\pi+\left(\frac{\pi}{6}+2 \alpha\right)\right)}{2 \cos \left(\frac{\pi}{6}-2 \alpha\right)+\sqrt{3} \cos (3 \pi-2 \alpha)}=\frac{-\sin 2 \alpha-2 \cos \left(\frac{\pi}{6}+2 \alpha\right)}{2 \cos \left(\frac{\pi}{6}-2 \alpha\right)-\sqrt{3} \cos 2 \alpha}= \\
& =\frac{\sin 2 \alpha+2 \cos \left(\frac{\pi}{6}+2 \alpha\right)}{\sqrt{3} \cos 2 \alpha-2 \cos \left(\frac{\pi}{6}-2 \alpha\right)}=[\cos (x \pm y)=\cos x \cos y \mp \sin x \sin y]= \\
& =\frac{\sin 2 \alpha+2 \cos \frac{\pi}{6} \cos 2 \alpha-2 \sin \frac{\pi}{6} \sin 2 \alpha}{\sqrt{3} \cos 2 \alpha-2 \cos \frac{\pi}{6} \cos 2 \alpha-2 \sin \frac{\pi}{6} \sin 2 \alpha}= \\
& =\frac{\sin 2 \alpha+2 \cdot \frac{\sqrt{3}}{2} \cos 2 \alpha-2 \cdot \frac{1}{2} \sin 2 \alpha}{\sqrt{3} \cos 2 \alpha-2 \cdot \frac{\sqrt{3}}{2} \cos 2 \alpha-2 \cdot \frac{1}{2} \sin 2 \alpha}= \\
& \frac{\sin 2 \alpha+\sqrt{3} \cos 2 \alpha-\sin 2 \alpha}{\sqrt{3} \cos 2 \alpha-\sqrt{3} \cos 2 \alpha-\sin 2 \alpha}=\frac{\sqrt{3} \cos 2 \alpha}{-\sin 2 \alpha}=-\sqrt{3} \operatorname{ctg} 2 \alpha . \\
& \text { Answer: }-\sqrt{3} \operatorname{ctg} 2 \alpha . \\
& 3.246 \cdot \frac{\cos 2 \alpha-\cos 6 \alpha+\cos 10 \alpha-\cos 14 \alpha}{\sin 2 \alpha+\sin 6 \alpha+\sin 10 \alpha+\sin 14 \alpha} . \\
& \text { Solution. } \\
& \frac{\cos 2 \alpha-\cos 6 \alpha+\cos 10 \alpha-\cos 14 \alpha}{\sin 2 \alpha+\sin 6 \alpha+\sin 10 \alpha+\sin 14 \alpha}=
\end{aligned}
$$

$$
\begin{aligned}
& =\frac{(\cos 10 \alpha+\cos 2 \alpha)-(\cos 14 \alpha+\cos 6 \alpha)}{(\sin 10 \alpha+\sin 2 \alpha)+(\sin 14 \alpha+\sin 6 \alpha)}= \\
& =\left[\cos x+\cos y=2 \cos \frac{x+y}{2} \cos \frac{x-y}{2}\right. \\
& \left.\sin x+\sin y=2 \sin \frac{x+y}{2} \cos \frac{x-y}{2}\right]= \\
& =\frac{2 \cos 6 \alpha \cos 4 \alpha-2 \cos 10 \alpha \cos 4 \alpha}{2 \sin 6 \alpha \cos 4 \alpha+2 \sin 10 \alpha \cos 4 \alpha}=\frac{2 \cos 4 \alpha(\cos 6 \alpha-\cos 10 \alpha)}{2 \cos 4 \alpha(\sin 6 \alpha+\sin 10 \alpha)}= \\
& =\frac{\cos 6 \alpha-\cos 10 \alpha}{\sin 6 \alpha+\sin 10 \alpha}=\left[\cos x-\cos y=-2 \sin \frac{x+y}{2} \sin \frac{x-y}{2}\right]= \\
& =\frac{-2 \sin 8 \alpha \sin (-2 \alpha)}{2 \sin 8 \alpha \cos 2 \alpha}=\operatorname{tg} 2 \alpha .
\end{aligned}
$$

Answer: $\operatorname{tg} 2 \alpha$.