2. The real solutions $(x, y)=$ of the equation $\sin (3 \cos x)-2 \sqrt{\sin (3 \cos x)} \cdot \sin (2 \cos y)+1=0$
2. $\left(2 k \pi \pm \arccos \frac{\pi}{6}, 2 k \pi \pm \arccos \frac{\pi}{4}, k \in Z\right)$. The original equation can be transformed as follows:
$$
\begin{array}{l}
{\left[\sqrt{\sin (3 \cos x)}-\sin (2 \cos y) j^{2}+\cos ^{2}(2 \cos y)=0\right.} \\
\Rightarrow\left\{\begin{array}{l}
\sqrt{\sin (\overline{\cos x})}=\sin (2 \cos y), \\
\cos (2 \cos y)=0 .
\end{array}\right. \\
\end{array}
$$
$$
\begin{array}{l}
\text { Then }\left\{\begin{array} { l } 
{ \operatorname { s i n } ( 2 \operatorname { c o s } y ) = 1 , } \\
{ \operatorname { s i n } ( 3 \operatorname { c o s } x ) = 1 , }
\end{array} \text { i.e., } \left\{\begin{array}{l}
2 \cos y=2 k \pi+\frac{\pi}{2}, \\
3 \cos x=2 k \pi+\frac{\pi}{2} .
\end{array}\right.\right. \\
\therefore\left\{\begin{array}{l}
\cos y=k \pi+\frac{\pi}{4}, \\
\cos x=\frac{2}{3} k \pi+\frac{\pi}{6} .
\end{array} \quad(k \in Z)\right. \\
\because|\cos x| \leqslant 1, \text { only } k=0, \\
\therefore x=2 k \pi \pm \arccos \frac{\pi}{6}, \\
y=2 k \pi \pm \arccos \frac{\pi}{4}(k \in Z) .
\end{array}
$$