3.141. $\sin 2 \alpha+\sin 4 \alpha+\sin 6 \alpha$.
Solution.

$$
\begin{aligned}
& \sin 2 \alpha+\sin 4 \alpha+\sin 6 \alpha .=(\sin 2 \alpha+\sin 4 \alpha)+\sin 2(3 \alpha)= \\
& =2 \sin 3 \alpha \cos \alpha+2 \sin 3 \alpha \cos 3 \alpha=2 \sin 3 \alpha(\cos \alpha+\cos 3 \alpha)= \\
& =2 \sin 3 \alpha \cdot 2 \cos 2 \alpha \cos \alpha=4 \sin 3 \alpha \cos 2 \alpha \cos \alpha
\end{aligned}
$$

Answer: $4 \sin 3 \alpha \cos 2 \alpha \cos \alpha$.