11.3. Find all values of the parameter $a$ for which the equation has a unique solution: $\boldsymbol{a}$ ) $\left.a x^{2}+\sin ^{2} x=a^{2}-a ; \boldsymbol{6}\right) a x^{2}+\sin ^{2} x=a^{3}-a$.
Answer: a) $a=1$; b) $a=1, a=-1$. Solution. a) See problem 10.3. b) As in the solution of problem 10.3, the question reduces to checking the values of $a$ for which $a^{3}-a=0$. For $a=0$ and $a=1$, the result has already been obtained in problem 10.3. For $a=-1$, we have the equation $\sin ^{2} x=x^{2} \Leftrightarrow|\sin x|=|x|$. However, as is known (for example, from the proof of the first remarkable limit), $|\sin x|<|x|$ for $x \neq 0$. Thus, the equation also has a unique solution for $a=-1$. Remark. The mentioned inequality can also be proven using the derivative, by considering the function $y=x-\sin x$ for non-negative $x$; this function has a non-negative derivative and is zero at $x=0$.