10.3. Find all values of the parameter $a$ for which the equation $a x^{2}+\sin ^{2} x=a^{2}-a$ has a unique solution.
Answer: $a=1$. Solution. Note that only $x=0$ can be the unique root of the equation, since due to the evenness of the functions involved, for any solution $x_{0} \neq 0$, $(-x_{0})$ will also be a solution. Therefore, we necessarily get $a^{2}-a=0 \Leftrightarrow a=0$ or $a=1$. Let's check these values. When $a=0$, we have the equation $\sin ^{2} x=0$, which has an infinite set of solutions $x=k \pi(k \in Z)$. If $a=1$, then the equation $x^{2}+\sin ^{2} x=0$ has a unique solution $x=0$.