1. Let $\triangle A B C$ be a given acute triangle, then the solution set of the equation $x^{2} \sin ^{2} B+x\left(\cos ^{2} C-\cos ^{2} A+\sin ^{2} B\right)$ $+\sin ^{2} A \cos ^{2} C=0$ with respect to $x$ is $\qquad$
1. Solution: By the cosine theorem, $a^{2}+b^{2}-c^{2}-2 a b \cos C=0$, and by the sine theorem, we have
$$
\begin{array}{c}
\sin ^{2} A+\sin ^{2} B-\sin ^{2} C-2 \sin A \sin B \cos C=0 \\
\Delta=\left(\sin ^{2} A+\sin ^{2} B-\sin ^{2} C\right)^{2}- \\
4 \sin ^{2} A \sin ^{2} B \cos ^{2} C \\
=\left(\sin ^{2} A+\sin ^{2} B-\sin ^{2} C+2 \sin A \sin B \cos C\right) \cdot \\
\left(\sin ^{2} A+\sin ^{2} B-\sin ^{2} C-2 \sin A \sin B \cos C\right)=0
\end{array}
$$

Therefore, the original equation has equal roots $x_{1}=x_{2}=-\frac{\sin A \cos C}{\sin B}$,
so the solution set is $\left\{-\frac{\sin A \cos C}{\sin B}\right\}$.