Question

Sinπx=1 cos(πx/4)=0 cos(πx/7)=-1 cos(πx/18)=-корень(3)/2 cos(πx/6)=-0,5

Answer 1

$1) \;\; sin(\pi x) = 1\\\\\pi x = \dfrac{\pi}2} + 2\pi n\\\\x = \dfrac{1}{2} + 2n, \;\;n \in Z\\\\\\2) \;\;cos(\frac{\pi x}{4}) = 0\\\\\dfrac{\pi x}{4} = \dfrac{\pi}{2} + \pi n\\\\x = \dfrac{\pi}{2}\cdot \dfrac{4}{\pi} + \pi n \cdot \dfrac{4}{\pi}\\\\x = 2 + 4n,\;\; n \in Z\\\\\\3)\;\; cos(\frac{\pi x}{7}) = -1 \\\\\dfrac{\pi x}{7} = \pi + 2\pi n\\\\x = \pi \cdot \frac{7}{\pi} + 2\pi n \cdot \frac{7}{\pi}\\\\x = 7 + 14n, n \in Z\\\\\\$

$4) \;\; cos(\frac{\pi x}{18})=-\frac{\sqrt3}{2}\\\\\frac{\pi x}{18}=\pm\underbrace{arccos\Big(-\frac{\sqrt3}{2}\Big)}_{\pi- arccos\frac{\sqrt3}{2}}+2\pi n\\\\\frac{\pi x}{18}=\pm (\pi-\frac{\pi}{6})+2\pi n\\\\\frac{\pi x}{18}=\pm\frac{5\pi}{6}+2\pi n\\\\x=\pm\frac{5\pi}{6}\cdot \frac{18}{\pi}+2\pi n\cdot\frac{18}{\pi}\\\\x=\pm15+36n,\;\;n\in Z$

$5)\;\;cos(\frac{\pi x}{6})=-\frac{1}{2}\\\\\frac{\pi x}{6}=\pm arccos(-\frac{1}{2})+2\pi n\\\\\frac{\pi x}{6}=\pm (\pi-\frac{\pi}{3})+2\pi n \\\\x = \pm \frac{2\pi}{3}\cdot \frac{6}{\pi} + 2\pi n \cdot \frac{6}{\pi}\\\\x = \pm 4 + 12n,\;\; n \in Z$

Z - это множество целых чисел