Question

16.Решить уравнение

Answer 1

$3 { sin}^{2} x + sinxcosx = 2cos ^{2} x \: \: \: |: {cos}^{2} x, \: cosx \neq 0 \\ \\ 3tg ^{2} x + tgx = 2 \\ 3tg ^{2} x + tgx - 2 = 0 \\ D = 1 + 4 \times 2 \times 3 = 1 + 24 = 25 = {5}^{2} \\1)tgx = \frac{ - 1 + 5}{2 \times 3} = \frac{4}{6} = \frac{2}{3} \\ \\ x = arctg \frac{2}{3} + \pi n, \: n \in Z \\ \\ 2)tgx = \frac{ - 1 - 5}{2 \times 3} = \frac{ - 6}{6} =- 1 \\ \\ x = - \frac{\pi}{4} + \pi n, \: n \in Z \\ \\ OTBET:arctg \frac{2}{3} + \pi n ; \: \frac{\pi}{4} + \pi n, \: n \in Z$