sin^2x-9sinxcosx+3cos^2x=-1

$sin^2x - 9sinxcosx + 3cos^2x = -1 \\ \\ sin^2x - 9sinxcosx + 3cos^2x = -sin^2x - cos^2x \\ \\ 2sin^2x - 9sinxcosx + 4cos^2x = 0 \ \ |:cos^2x \\ \\ 2tg^2x - 9tgx + 4 = 0$

Пусть $t = tgx.$

$2t^2 - 9t + 4 = 0 \\ \\ D = 81 - 4 \cdot 4 \cdot 2 = 81 - 32 = 49 = 7^2 \\ \\ t_1 = \dfrac{9+7}{4} = \dfrac{16}{4} = 4 \\ \\ t_2 = \dfrac{9-7}{4} = \dfrac{1}{2}$

Обратная замена:

$tgx = 4 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ tgx = \dfrac{1}{2} \\ \\ x = arctg4 + \pi n, \ n \in Z \ \ \\ and \\ x = arctg\dfrac{1}{2} + \pi k, \ k \in Z$