11.1. Solve the equation $(\sin 2 x-\pi \sin x) \sqrt{11 x^{2}-x^{4}-10}=0$.
Answer: $x \in\{-\sqrt{10},-\pi,-1,1, \pi, \sqrt{10}\}$. Solution. Let's find the domain. We have $11 x^{2}-x^{4}-10 \geq 0 \Leftrightarrow\left(x^{2}-10\right)\left(x^{2}-1\right) \leq 0 \Leftrightarrow-\sqrt{10} \leq x \leq-1$ or $1 \leq x \leq \sqrt{10}$. In this case, we obtain four values $x \in\{-\sqrt{10},-1,1, \sqrt{10}\}$, for which the expression under the square root turns to zero. The first factor of the equation $\sin 2 x-\pi \sin x=2 \sin x\left(\cos x-\frac{\pi}{2}\right)$ turns to zero only under the condition $\sin x=0$ (since $\pi / 2>1$). Thus, $x=k \pi(k \in Z)$, but only two values from this series fall into the domain, namely $\pm \pi$ (the inequality $\pi<\sqrt{10}$ follows from the inequalities $\pi<3.15$ and $\left.(3.15)^{2}<10\right)$.