Решить уравнение!!!!

$3 {sin}^{2} x + sinx \times cosx = 2 {cos}^{2} x \\ 3 {sin}^{2} x + sinx \times cosx - 2 {cos}^{2} x = 0 \\ \frac{3 {sin}^{2} x}{ {cos}^{2}x } + \frac{sinx \times cosx}{ {cos}^{2}x } - \frac{2 {cos}^{2} x}{ {cos}^{2}x } = 0 \\ 3 {tg}^{2} x + tgx - 2 = 0 \\ tgx = t \\ 3 {t}^{2} + t - 2 = 0 \\ d = {b}^{2} - 4ac = 1 - 4 \times 3 \times ( - 2) = 25 \\ x1 = \frac{ - 1 + 5}{2 \times 3} = \frac{4}{6} = \frac{2}{3} \\ x2 = \frac{ - 1 - 5}{2 \times 3} = \frac{ - 6}{6} = - 1 \\ 1)tgx = \frac{2}{3} \\ x = arctg \frac{2}{3} + \pi n \\ 2)tgx = - 1 \\ x = arctg( - 1) + \pi n \\ x = - \frac{\pi}{4} + \pi n$
Ответ:
arctg2/3 + pi*n, n € Z;
-pi/4 + pi*n, n € Z.