1.1. Find the greatest negative root of the equation $\sin 2 \pi x=\sqrt{3} \sin \pi x$.
Solution. The given equation is equivalent to

$$
\sin \pi x(2 \cos \pi x-\sqrt{3})=0 \Longleftrightarrow\left[\begin{array} { l } 
{ \pi x = \pi n , n \in \mathbb { Z } } \\
{ \pi x = \pm \frac { \pi } { 6 } + 2 \pi k , k \in \mathbb { Z } , }
\end{array} \Longleftrightarrow \left[\begin{array}{l}
x=n, n \in \mathbb{Z} \\
x= \pm \frac{1}{6}+2 k, k \in \mathbb{Z}
\end{array}\right.\right.
$$

Therefore, the greatest negative root of the equation is $-\frac{1}{6}$.

Answer: $-\frac{1}{6}$. $(\mathrm{E})$

$$
\mathbf{A}-1 \quad \mathbf{B}-\frac{5}{6} \quad \mathbf{C}-\frac{2}{3} \quad \mathbf{D}-\frac{1}{3} \quad \mathbf{E}-\frac{1}{6} \quad \mathbf{F}
$$