Question

Решите пожалуйста.....

Answer 1

Ответ:

3.

Пусть cosx = t

$2 {t}^{2} - 3t + 1 = 0 \\ d = 9 - 8 = 1 \\ t1 = \frac{3 + 1}{4} = 1 \\ t2 = \frac{1}{2} \\ \\ \cos(x) = 1 \\ x1 = 2\pi \: n \\ \\ \cos(x) = \frac{1}{2} \\ x2 = + - \frac{\pi}{3} + 2 \pi \: n$

4.

разделим на cos^2x, не равный 0

$3 {tg}^{2} x + tgx - 2 = 0 \\ \\ tgx = t \\ 3 {t}^{2} + t - 2 = 0 \\ d = 1 + 24 = 25 \\ t1 = \frac{ - 1 + 5}{6} = \frac{2}{3} \\ t2 = - 1 \\ \\ tgx = \frac{2}{3} \\ x1 = arctg( \frac{2}{3} ) + \pi \: n \\ \\ tgx = - 1 \\ x2 = - \frac{\pi}{4} + \pi \: n$

n принадлежит Z.