1. It is known that $\sin y=\frac{3}{2} \sin x+\frac{2}{3} \cos x, \cos y=\frac{2}{3} \sin x+\frac{3}{2} \cos x$. Find $\sin 2 x$.
Answer. $\sin 2 x=-\frac{61}{72}$.

Solution. Squaring both equalities and adding them term by term, we get $1=\frac{9}{4}+4 \sin x \cos x+\frac{4}{9}$, from which $-2 \sin 2 x=\frac{61}{36}, \sin 2 x=-\frac{61}{72}$.