Question

Найти все корни уравнения cos 2x= 1/2, на отрезке [ -п/2; 5п/2 ] 

Срочно нужно. Заранее спасибо)

Answer 1

$\cos 2x =\frac{1}{2}\;\;\;\;\;x\in\big[-\frac{\big\pi}{2}, \frac{5\big\pi}{2}\big]\\\\2x = \pm\frac{\big\pi}{3} + 2\pi n,\; n \in \mathbb Z\\\\x = \pm\frac{\big\pi}{6} + \pi n,\; n \in \mathbb Z$

$1.\;\; x = \frac{\big\pi}{6} + \pi n,\; n \in \mathbb Z \;\;\;\;x\in\big[-\frac{\big\pi}{2}, \frac{5\big\pi}{2}\big]\\\\n = -1; \;\; x = \frac{\big\pi}{6} - \pi = -\frac{5\big\pi}{6} \notin\big[-\frac{\big\pi}{2},\frac{5\big\pi}{2}\big]\\\\n = 0;\;\; x = \frac{\big\pi}{6}\\\\n = 1;\;\; x = \frac{7\big\pi}{6}\\\\n = 2;\;\; x = \frac{13\big\pi}{6}$

$2.\;\;x = -\frac{\big\pi}{6} + \pi n,\; n \in \mathbb Z \;\;\;\;x\in\big[-\frac{\big\pi}{2}, \frac{5\big\pi}{2}\big]\\\\n = 0;\;\;x = -\frac{\big\pi}{6}\\\\n = 1;\;\; x = \frac{5\big\pi}{6}\\ \\ n = 2;\;\; x = \frac{11\big\pi}{6}$

Ответ: ± π/6, 5π/6, 7π/6, 11π/6, 13π/6