Question

35 баллов за задание
А77

Answer 1

$\displaystyle\bf\\Sin2\pi xCos3\pi x-Sin4\pi xCos\pi x=0\\\\\\\frac{1}{2} \Big[Sin(2\pi x-3\pi x)+Sin(2\pi x+3\pi x)\Big]-\frac{1}{2} \Big[Sin(4\pi x-\pi x)+Sin(4\pi x+\pi x)\Big]=0\\\\\\-Sin\pi x+Sin5\pi x-Sin3\pi x-Sin5\pi x=0\\\\\\Sin3\pi x+Sin\pi x=0\\\\\\2Sin\frac{3\pi x+\pi x}{2} Cos\frac{3\pi x-\pi x}{2} =0\\\\\\Sin2\pi x Cos\pi x=0$

$\displaystyle\bf\\\left[\begin{array}{ccc}Sin2\pi x=0\\Cos\pi x=0\end{array}\right\\\\\\\left[\begin{array}{ccc}2\pi x=\pi n,n\in Z\\\pi x=\frac{\pi }{2} +\pi n,n\in Z\end{array}\right\\\\\\\left[\begin{array}{ccc}x=\dfrac{n}{2},n\in Z \\x=\dfrac{1}{2} +n,n\in Z\end{array}\right\\\\\\1)\\x=\frac{n}{2} \\\\n=0 \ \ \ \Rightarrow \ \ x=0\\\\n=1 \ \ ; \ \ \Rightarrow \ \ x=\frac{1}{2} \\\\n=2 \ \ ; \ \ \Rightarrow \ \ x=1\\\\2)\\x=\frac{1}{2} +n$

$\displaystyle\bf\\ n=0 \ \ ; \ \ \Rightarrow \ \ x=\frac{1}{2} \\\\Otvet:3$