11.1. How many roots does the equation $\sqrt{14-x^{2}}(\sin x-\cos 2 x)=0$ have?
Answer: 6 roots. Solution. The domain of the equation: $14-x^{2} \geq 0 \Leftrightarrow|x| \leq \sqrt{14}$. On this domain, we solve the trigonometric equation $\sin x-\cos 2 x=0 \Leftrightarrow \sin x+2 \sin ^{2} x-1=0$; $\sin x=\frac{-1 \pm 3}{4}$, i.e., $\sin x=-1$ or $\sin x=\frac{1}{2}$. From this, we get three series of solutions: 1) $\left.\left.x=-\frac{\pi}{2}+2 \pi k ; 2\right) x=\frac{\pi}{6}+2 \pi n ; 3\right) x=\frac{5 \pi}{6}+2 \pi m(k, n, m \in Z)$. Solutions of the first series lie in the domain of definition only when $k=0$ (since for $k \neq 0,|x| \geq 2 \pi-\frac{\pi}{2}=\frac{3 \pi}{2}>\frac{9}{2}>\sqrt{14}$ ). Solutions of the second series also lie in the domain of definition only when $n=0$ (since $\left|\frac{\pi}{6}-2 \pi\right|=\frac{11 \pi}{6}>\frac{11}{2}>\sqrt{14}$ ). Solutions of the third series lie in the domain of definition for $m=0$ and $m=-1$. Indeed, $\frac{5}{6} \pi2 \pi>\sqrt{14}$ ). Adding the roots of the first factor $\sqrt{14-x^{2}}$ of the original equation, namely $\pm \sqrt{14}$, we get a total of 6 roots.