3.265. $\sin (2 x-\pi) \cos (x-3 \pi)+\sin \left(2 x-\frac{9 \pi}{2}\right) \cos \left(x+\frac{\pi}{2}\right)$.
## Solution.

$$
\begin{aligned}
& \sin (2 x-\pi) \cos (x-3 \pi)+\sin \left(2 x-\frac{9 \pi}{2}\right) \cos \left(x+\frac{\pi}{2}\right)= \\
& =-\sin (\pi-2 x) \cos (3 \pi-x)-\sin \left(4 \pi+\left(\frac{\pi}{2}-2 x\right)\right) \cos \left(\frac{\pi}{2}+x\right)= \\
& =-\sin 2 x(-\cos x)-\cos 2 x(-\sin x)=\sin 2 x \cos x+\cos 2 x \sin x= \\
& =[\sin \alpha \cos \beta+\cos \alpha \sin \beta=\sin (\alpha+\beta)]=\sin 3 x
\end{aligned}
$$

Answer: $\sin 3 x$.