6.9. (GDR, 66). Solve the equation

$$
(\sin (x-y)+1)(2 \cos (2 x-y)+1)=6
$$
6.9. Due to the estimates $-1 \leqslant \sin \alpha \leqslant 1, -1 \leqslant \cos \alpha \leqslant 1$, we have the inequality

$$
(\sin (x-y)+1)(2 \cos (2 x-y)+1)<6
$$

in which equality is possible only in the case when

$$
\sin (x-y)=1 \quad \text { and } \quad \cos (2 x-y)=1 .
$$

Therefore, the equation is equivalent to the system

$$
\left\{\begin{aligned}
\sin (x-y) & =1 \\
\cos (2 x-y) & =1,
\end{aligned}\right.
$$

i.e., the system

$$
\left\{\begin{aligned}
x-y & =\frac{\pi}{2}+2 \pi m \quad(m \in \mathbb{Z}) \\
2 x-y & =2 \pi n \quad(n \in \mathbb{Z})
\end{aligned}\right.
$$

Thus, we have

$$
\begin{gathered}
x=-\frac{\pi}{2}+2(n-m) \pi=2 \pi k-\frac{\pi}{2} \quad(k \in \mathbb{Z}) \\
y=(2(n-2 m)-1) \pi=(2(k-m)-1) \pi=(2 l+1) \pi \quad(l \in \mathbb{Z})
\end{gathered}
$$