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

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

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