Найти все корни уравнения [latex]cos2x=- \frac{1}{2} [/latex] на отрезке [latex][- \frac{ \pi }{2}; \frac{5 \pi }{2} ][/latex] получается: х=+- [latex] \frac{ \pi }{3} + \pi n[/latex] я нашел только x=[latex] \frac{ \pi }{3} [/...

[latex]\cos 2x = -\frac{1}{2} \\ \\ 2x = \pm \frac{2\pi}{3} + 2 \pi n, \ n \in Z \\ \\ x = \pm \frac{\pi}{3}+ \pi n, \ n \in Z \\ \\ -\frac{\pi}{2} \leq \frac{\pi}{3}+ \pi n \leq \frac{5\pi}{2}; \ \ \ \ \ \ \ \ \ -\frac{\pi}{2} \leq -\frac{\pi}{3}+ \pi k \leq \frac{5\pi}{2} \\ \\ -\frac{\pi}{2} - \frac{\pi}{3}\leq \pi n \leq \frac{5\pi}{2} - \frac{\pi}{3}; \ \ \ \ -\frac{\pi}{2} + \frac{\pi}{3}\leq \pi k \leq \frac{5\pi}{2} + \frac{\pi}{3} \\ \\ [/latex] [latex]\\ \\ -\frac{5}{6} \leq n \leq \frac{13}{6} ; \ \ \ \ \ \ \ \ \ -\frac{1}{6} \leq k \leq \frac{17}{6} \\ \\ n=0, \ 1, \ 2; \ \ \ \ \ \ \ \ \ \ \ \ \ k=0, \ 1, \ 2 \\ \\ \\ 1) \ \frac{\pi}{3}; \ 2) \ \frac{\pi}{3}+ \pi =\frac{4 \pi}{3}; \ 3) \ \frac{\pi}{3}+ 2\pi =\frac{7 \pi}{3} \\ \\ 4) \ -\frac{\pi}{3}; \ 5) \ -\frac{\pi}{3}+ \pi = \frac{2 \pi}{3}; \ 6) \ -\frac{\pi}{3}+ 2\pi =\frac{5\pi}{3}[/latex]