Question

Реши уравнение cos(2π+8x)=\sqrt{2}/2и выбери корни из отрезка [π/2;π].

Answer 1

$\cos(2\pi + 8x) = \frac{ \sqrt{2} }{2} \\ \cos(8x) = \frac{ \sqrt{2} }{2} \\ \left[ \begin{gathered} 8x = \frac{\pi}{4} + 2\pi n \\8x = - \frac{\pi}{4} + 2\pi n \end{gathered} \right. \\ \left[ \begin{gathered} x = \frac{\pi}{32} + \frac{\pi}{4}n,\: n \in \mathbb Z \\ x = - \frac{\pi}{32} + \frac{\pi}{4} m, \: m \in \mathbb Z \end{gathered} \right.$

Отбираем корни:

$\frac{\pi}{2} \leqslant \frac{\pi}{32} + \frac{\pi}{4} n \leqslant \pi \\ \frac{1}{2} - \frac{1}{32} \leqslant \frac{1}{4} n \leqslant 1 - \frac{1}{32} \\ \frac{15 \times 4}{32} \leqslant n \leqslant \frac{31 \times 4}{32} \\ \frac{15}{8} \leqslant n \leqslant \frac{31}{8} \\ n = \{1,2,3 \} \\ \left[ \begin{gathered}x_{1} = \frac{\pi}{32} + \frac{\pi}{4} = \frac{9\pi}{32} \\ x_{2} = \frac{\pi}{32} + \frac{2\pi}{4} = \frac{17\pi}{32} \\ x_{3} = \frac{\pi}{32} + \frac{3\pi}{4} = \frac{25\pi}{32} \end{gathered} \right.$

$\frac{\pi}{2} \leqslant - \frac{\pi}{32} + \frac{\pi}{4} m \leqslant \pi \\ \frac{1}{2} + \frac{1}{32} \leqslant \frac{1}{4} m \leqslant 1 + \frac{1}{32} \\ \frac{17 \times 4}{32} \leqslant m \leqslant \frac{33 \times 4}{32} \\ \frac{17}{8} \leqslant m \leqslant \frac{33}{8} \\ m = \{2,3,4 \} \\ \left[ \begin{gathered}x_{1} = - \frac{\pi}{32} + \frac{\pi}{2} = \frac{15\pi}{32} \\ x_{2} = - \frac{\pi}{32} + \frac{3\pi}{4} = \frac{23\pi}{32} \\ x_{3} = - \frac{\pi}{32} + \frac{4\pi}{4} = \frac{31\pi}{32} \end{gathered} \right.$