Cos5x>=-1/2 решите неравенство

Cos 5x ≥ 1/2

Находим нули функции:

cos 5x = - 1/2

5x = ± arccos (-1/2) + 2πK, K ϵ Z

5x = ± (π - arccos1/2) + 2πK, K ϵ Z

5x = ± (π - π/3) + 2πK, K ϵ Z

5x = ± 2π/3 + 2πK, K ϵ Z

x = ± 2π/15 + 2π/5 * K, K ϵ Z.

Так как сos 5x ≥ 1/2, то получаем:

- 2π/15 + 2π/5 * K ≤ х ≤ 2π/15 + 2π/5 * K, K ϵ Z.

Ответ: х ϵ [ - 2π/15 + 2π/5 * K; 2π/15 + 2π/5 * K], K ϵ Z.