11.2. Solve the equation $\sqrt{-x^{2}+x+2} \cdot(\sin 2 x-\pi \cos x)=0$.
Answer. $x_{1}=-1 ; x_{2}=2 ; x_{3}=\frac{\pi}{2}$. Solution. The expression under the root $-x^{2}+x+2$ gives two roots $x_{1}=-1 ; x_{2}=2$ and defines the domain of definition: $-1 \leq x \leq 2$. By equating the bracket $\sin 2 x-\pi \cos x=$ $\cos x(2 \sin x-\pi)$ to zero, we obtain the system of equations $\left[\begin{array}{l}\cos x=0 \\ \sin x=\frac{\pi}{2}\end{array}\right.$. The first equation of the system, taking into account the domain of definition, gives $x=\frac{\pi}{2}$, while the second has no solutions, since $\frac{\pi}{2}>1$.