3.324. $\sin 2 \alpha+\cos 4 \alpha-\sin 6 \alpha$.
Solution.

$\sin 2 \alpha+\cos 4 \alpha-\sin 6 \alpha=(\sin 2 \alpha-\sin 6 \alpha)+\cos 4 \alpha=$ $=\left[\sin x-\sin y=2 \cos \frac{x+y}{2} \sin \frac{x-y}{2}\right]=2 \cos 4 \alpha \sin (-2 \alpha)+\cos 4 \alpha=$ $=-2 \cos 4 \alpha \sin 2 \alpha+\cos 4 \alpha=-2 \cos 4 \alpha\left(\sin 2 \alpha-\frac{1}{2}\right)=$

$-2 \cos 4 \alpha\left(\sin 2 \alpha-\sin 30^{\circ}\right)=-2 \cos 4 \alpha \cdot 2 \cos \left(\alpha+15^{\circ}\right) \sin \left(\alpha-15^{\circ}\right)=$ $=4 \cos 4 \alpha \sin \left(15^{\circ}-\alpha\right) \cos \left(15^{\circ}+\alpha\right)$.

Answer: $4 \cos 4 \alpha \sin \left(15^{\circ}-\alpha\right) \cos \left(15^{\circ}+\alpha\right)$.