1. Solution. The equation is equivalent to $\sin (\pi / 4-11 x)=\sin (17 x+\pi / 4)$. From this, either $\pi / 4-11 x=17 x+\pi / 4+2 \pi k, k \in \mathbb{Z}$, or $\pi-(\pi / 4-11 x)=$ $17 x+\pi / 4+2 \pi n, n \in \mathbb{Z}$.
Answer: $\pi k / 14, \pi / 12+\pi n / 3, k, n \in \mathbb{Z}$ (in the variant $5-2: \pi k / 17,-\pi / 12+\pi n / 3$, $k, n \in \mathbb{Z})$.