Question

СРОЧНО ПОМОГИТЕ, ПРОФИЛЬНАЯ МАТЕМАТИКА ПРОШУ

Answer 1

Ответ:

$4 \cos {}^{4} (x) - 4\cos {}^{2} (x) + 3\sin(x) \cos(x) + 1 = 0 \\ 4 \cos {}^{2} (x)( \cos {}^{2} (x) - 1) + \frac{3}{2} \times 2 \sin(x) \cos(x) + 1 = 0 \\ 4 \cos {}^{2} (x) \times ( - \sin {}^{2} (x)) + \frac{3}{2} \sin(2x) + 1 = 0 \\ - \sin {}^{2} (2x) + 1.5 \sin(2x) +1 = 0 \: \: \: | \times ( - 2) \\ 2 \sin {}^{2} (2x) - 3 \sin(2x) - 2 = 0 \\ \\ \sin(2x) = t \\ \\ 2t {}^{2} - 3 t - 2 = 0\\ D =9 + 16 = 25 \\ t_1 = \frac{3 + 5}{4} = 2 \\ t_2 = - \frac{1}{2} \\ \\ \sin(2x) = 2$

нет корней

$\sin(2x) = - \frac{1}{2} \\ \\ 2x_1 = - \frac{\pi}{6} + 2\pi \: n \\ x_1 = - \frac{\pi}{12} + \pi \: n \\ \\ 2x_2 = - \frac{5\pi}{6} + 2\pi \: n \\ x_2 = - \frac{5\pi}{12} + \pi \: n$

n принадлежит Z.

На промежутке:

рисунок:

$x_1 = 3\pi - \frac{5\pi}{12} = \frac{36 - 5}{12} \pi = \frac{31\pi}{12} \\ x_2 = 3\pi - \frac{\pi}{12} = \frac{35\pi}{12}$

Ответ:

$1)x_1 = - \frac{\pi}{12} + \pi \: n \\ x_2 = - \frac{5\pi}{12} + \pi \: n \\ 2) \frac{31\pi}{12} ;\frac{35\pi}{12}$