3.285. $\sin 6 \alpha-2 \sqrt{3} \cos ^{2} 3 \alpha+\sqrt{3}$.
Solution.

$$
\begin{aligned}
& \sin 6 \alpha-2 \sqrt{3} \cos ^{2} 3 \alpha+\sqrt{3}=\left[\cos ^{2} \frac{x}{2}=\frac{1+\cos x}{2}\right]= \\
& =\sin 6 \alpha-\sqrt{3} \cos 6 \alpha=2\left(\frac{1}{2} \sin 6 \alpha-\frac{\sqrt{3}}{2} \cos 6 \alpha\right)= \\
& =2\left(\cos 60^{\circ} \sin 6 \alpha-\sin 60^{\circ} \cos 6 \alpha\right)=2\left(\sin 6 \alpha \cos 60^{\circ}-\cos 6 \alpha \sin 60^{\circ}\right)= \\
& =[\sin x \cos y-\cos x \sin y=\sin (x-y)]=2 \sin \left(6 \alpha-60^{\circ}\right)
\end{aligned}
$$

Answer: $2 \sin \left(6 \alpha-60^{\circ}\right)$.