Найдите сумму корней уравнения sin*пи*x+ cos*пи*x=1 , принадлежащих отрезку [-3;1].

[latex]sin\pi x+cos\pi x=1\\\\\frac{1}{\sqrt2}\cdot sin\pi x+\frac{1}{\sqrt2}\cdot cos\pi x=\frac{1}{\sqrt2}\\\\cos\frac{\pi}{4}\cdot sin\pi x+sin\frac{\pi}{4}\cdot cos\pi x=\frac{1}{\sqrt2}\\\\sin(\pi x +\frac{\pi}{4})=\frac{1}{\sqrt2}\\\\\pi x+\frac{\pi}{4}=(-1)^{n}\cdot \frac{\pi}{4}+\pi n= \left \{ {{\frac{\pi}{4}+2\pi n,n\in Z} \atop {\frac{3\pi}{4}+2\pi k,k\in Z}} \right. [/latex] [latex]\pi x= \left \{ {{2\pi n,n\in Z} \atop {\frac{\pi}{2} +2\pi k,k\in Z}} \right. \\\\x= \left \{ {{2n,n\in Z} \atop {\frac{1}{2}+2k,k\in Z}} \right. [/latex] [latex]x\in [-3,1]:\; \; \frac{-3}{2},0,\frac{1}{2},\frac{5}{2}.[/latex] Сумма  корней = 3\2.