Question

Найти все корни уравнения cos4x=корень 2/2,удовлетворяющие неравенству модуль x<п/2

Answer 1

$|x| < \frac{\big\pi}{2}\;\;\Leftrightarrow\;\;-\frac{\big\pi}{2} < x < \frac{\big\pi}{2} \\\\ \cos 4x = \frac{\sqrt2}{2}\;\;\;\;\;\;x\in(-\frac{\big\pi}{2},\frac{\big\pi}{2})\\\\4x = \pm arccos(\frac{\sqrt2}{2}) + 2\pi n\\\\4x = \pm \frac{\big\pi}{4} + 2\pi n\\\\x = \pm\frac{\big\pi}{16} + \frac{\big\pi}{2} n, n \in \mathbb Z$

$1. \;\;x = \frac{\big\pi}{16} + \frac{\big\pi}{2} n, \;n\in\mathbb{Z}\;\;\;\;x\in(-\frac{\big\pi}{2},\frac{\big\pi}{2})\\\\-\frac{\big\pi}{2} < \frac{\big\pi}{16} + \frac{\big\pi}{2}n <\frac{\big\pi}{2}\;\;\;\Big| \times\ \frac{2}{\big\pi}\\\\-1 < \frac{1} {8} + n < 1\\\\-1\frac{1}{8} < n < \frac{7}{8}\\\\n = -1;\;\;\;x = \frac{\big\pi}{16} - \frac{\big\pi}{2} = \frac{\big\pi}{16} - \frac{8\big\pi}{16} = -\frac{7\big\pi}{16}\\\\n = 0;\;\;\; x = \frac{\big\pi}{16}$

$2.\;\;\; x = -\frac{\big\pi}{16} + \frac{\big\pi}{2} m, m\in\mathbb{Z}\;\;\;\;x\in(-\frac{\big\pi}{2},\frac{\big\pi}{2})\\\\-\frac{\big\pi}{2}< -\frac{\big\pi}{16} + \frac{\big\pi}{2} m<\frac{\big\pi}{2}\;\;\Big|\times\frac{2}{\big\pi}\\\\-1< -\frac{1} {8} + m < 1\\\\-\frac{7}{8} < m < 1\frac{1}{8}\\\\m = 0;\;\;\; x = -\frac{\big\pi}{16}\\\\m = 1;\;\;\; x = -\frac{\big\pi}{16} + \frac{\big\pi}{2} = -\frac{\big\pi}{16} + \frac{8\big\pi}{16} = \frac{7\big\pi}{16}$

Ответ:

$\frac{7\big\pi}{16}; -\frac{\big\pi}{16}; \frac{\big\pi}{16}; -\frac{7\big\pi}{16}$