Question

Решите уравнение прошу

Answer 1

3tg²x + 2tgx - 1 = 0
Сделаем замену   tgx = m
3m² + 2m - 1 = 0
D = 2² - 4 * 3 * (- 1) = 4 + 12 = 16 = 4²
$m_{1}= \frac{-2+4}{6} = \frac{1}{3} \\\\ m_{2} = \frac{-2-4}{6}=-1\\\\tgx= \frac{1}{3} \\\\x=arctg \frac{1}{3}+ \pi n \\\\\\tgx=-1\\\\x=arctg(-1)+ \pi n\\\\x=- \frac{ \pi }{4} + \pi n,n$∈z