Question

Решите пожалуйста подробно.

Answer 1

Ответ:

$2 \cos(x) - \cos(2x) - \cos {}^{2} (x) = 0 \\ 2 \cos(x) - (2 \cos {}^{2} (x) - 1) - \cos {}^{2} (x) = 0 \\ 2 \cos(x) - 2 \cos {}^{2} (x) + 1 - \cos {}^{2} (x) = 0 \\ - 3 \cos {}^{2} (x) + 2 \cos(x) + 1 = 0 \\ 3 \cos {}^{2} (x) - 2 \cos(x) - 1 = 0 \\ \\ \cos(x) = t \\ \\ 3t {}^{2} - 2t - 1 = 0\\ D = 4 + 12 = 16\\ t_1 = \frac{2 + 4}{6} = 1 \\ t_2 = - \frac{1}{3} \\ \\ \cos(x) = 1 \\ x_1 = 2\pi \: n \\ \\ \cos(x) = - \frac{1}{3} \\ x_2 = \pi\pm \: arccos( \frac{1}{3} ) + \pi \: n$

n принадлежит Z.