Question

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

Answer 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:

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