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

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