Question

Тригнометрия. Помогите решить, пожалуйста

Answer 1

Ответ:

$9 \sin(x) \cos( x ) - 7\cos {}^{2} (x) =2 \sin {}^{2} (x) \\ | \div \cos {}^{2} (x) \ne0 \\ 9tgx - 7 = 2tg {}^{2} x \\ 2 {tg}^{2} x - 9tgx + 7 = 0 \\ \\ tgx = t \\ \\ 2t {}^{2} - 9t + 7 = 0 \\ D = 81 - 56 = 25 \\ t_1 = \frac{9 + 5}{4} = \frac{7}{2} = 3.5 \\ t_2 = 1 \\ \\ tgx = 3.5 \\ x_1 = arctg(3.5) + \pi \: n \\ \\ tgx = 1 \\ x_2 = \frac{\pi}{4} + \pi \: n$

n принадлежит Z.

на промежутке:

$\\ x1 = - 3\pi + \frac{\pi}{4} = - \frac{11\pi}{4} \\ x2 = - 3\pi + arctg(3.5) \\ x3 = - 2\pi + \frac{\pi}{4} = - \frac{7\pi}{4} \\ x4 = - 2\pi + arctg(3.5)$