Найдите решение уравнения cos x/2 = 1/2 на отрезке от [ 0; 4пи]

cos(x/2)=1/2 ⇒ x/2=+-π/3+2πn, n∈Z ⇒ x=+-2π/3+4πn, n∈Z На отрезке x∈[0;4π] находятся 2 корня: 2π/3 и 10π/3 Ответ: 2π/3 и 10π/3

cos[latex] \frac{x}{2} = \frac{1}{2} [/latex] [latex] \frac{x}{2} = +-\frac{ \pi }{3} +2 \pi n[/latex] x=[latex]+- \frac{2 \pi }{3} +4 \pi n[/latex] [latex][0;4 \pi [/latex]] 1) [latex]0 \leq -\frac{2 \pi }{3} +4 \pi n \leq 4 \pi [/latex] [latex]0 \leq - \frac{2}{3} +4n \leq 4[/latex] [latex] \frac{2}{3} \leq 4 \pi \leq \frac{14}{3} [/latex] [latex] \frac{2}{12} \leq n \leq \frac{14}{12} [/latex] n=1; x=-[latex] \frac{2 \pi }{3} +4 \pi = \frac{10 \pi }{3} [/latex] 2)  [latex]0 \leq \frac{2 \pi }{3} +4 \pi n \leq 4 \pi [/latex] [latex]0 \leq \frac{2}{3} +4n \leq 4[/latex] [latex]- \frac{2}{3} \leq 4n \leq \frac{10}{3} [/latex] [latex]- \frac{2}{12} \leq n \leq \frac{10}{12} [/latex] n=0;x=[latex] \frac{2 \pi }{3} [/latex]