Question

Нужно решить неравенство:

Answer 1

Ответ:

$\frac{1}{ \cos {}^{2} (x) } + 3tgx - 5 = 0 \\ 1 + {tg}^{2} x + 3tgx - 5 = 0 \\ {tg}^{2} x + 3tgx - 4 = 0 \\ \\ tgx = t \\ \\ t {}^{2} + 3t - 4 = 0 \\ D= 9 + 16 = 25 \\ t_1 = \frac{ - 3 + 5}{2} = 1 \\ t_2 = - 4 \\ \\ tgx = 1 \\ x_1 = \frac{\pi}{4} + \pi \: n \\ \\ tgx = - 4 \\ x_2 = - arctg4 + \pi \: n \\ \\ n\in \: z$