7.1.1. (12 points) Find the greatest negative root of the equation

$$
\frac{\sin \pi x - \cos 2 \pi x}{(\sin \pi x - 1)^2 + \cos^2 \pi x - 1} = 0
$$
Answer: -0.5.

Solution. Transforming the equation, we get

$$
\frac{\sin \pi x-1+2 \sin ^{2} \pi x}{2 \sin \pi x-1}=0, \quad \frac{(\sin \pi x+1)(2 \sin \pi x-1)}{2 \sin \pi x-1}=0
$$

from which $\sin \pi x=-1, x=-\frac{1}{2}+2 n, n \in \mathbb{Z}$. The greatest negative number among these is $-\frac{1}{2}$.