Корень2cos(2x-П/5)-1=0

[latex]2cos(2x-\frac{\pi}{5})-1=0\\cos(2x-\frac{\pi}{5})=\frac{1}{2}\\2x-\frac{\pi}{5}=\pm \frac{\pi}{3}+2\pi n\\2x=\pm \frac{\pi}{3}+\frac{\pi}{5}+2\pi n\\x=\pm \frac{\pi}{6}+\frac{\pi}{10}+\pi n, \; n\in Z.[/latex] Можно ответ оставить таким либо записать корни по отдельности: [latex]2x-\frac{\pi}{5}= \frac{\pi}{3}+2\pi n\\2x=\frac{8\pi}{15}+2\pi n\\x= \frac{4\pi}{15}+\pi n, \; n\in Z;\\\\ 2x-\frac{\pi}{5}=- \frac{\pi}{3}+2\pi n\\2x=- \frac{2\pi}{15}+2\pi n\\x=- \frac{\pi}{15}+\pi n, \; n\in Z.[/latex]