1.2. Find the greatest negative root of the equation $\sin \pi x = -\sqrt{3} \cos \frac{\pi x}{2}$.
Solution. The roots of the equation are the following series of values: $x=1+2 n, x=-\frac{2}{3}+4 n, x=$ $-\frac{4}{3}+4 n, n \in \mathbb{Z}$. Therefore, the greatest negative root of the equation is $-\frac{2}{3}$.

Answer: $-\frac{2}{3} \cdot(\mathrm{D})$

$$
\mathbf{A}-\frac{4}{3} \quad \mathbf{B}-1 \quad \boxed{\mathbf{C}}-\frac{5}{6} \quad \mathbf{D}-\frac{2}{3} \quad \mathbf{E}-\frac{1}{6} \quad \mathbf{F}
$$