Question

3sin^2x-7sinxcosx+4cos^2x=0

Answer 1

$3sin^2x - 7sinxcosx + 4cos^2x = 0$
Разделим на $cos^2x$.
$3tg^2x - 7tgx + 4 = 0$
Пусть $t = tgx.$
$3t^2 - 7t + 4 = 0 \ D = 49 - 4 cdot 4 cdot 3 = 1 \ t_1 = frac{7 + 1}{6} = frac{4}{3} \ t_2 = frac{7 - 1}{6} = 1$
Обратная замена:
$tgx = frac{4}{3} \ x = boxed{arctg frac{4}{3} + pi n, n in Z} \ tgx = 1 \ boxed{x = frac{ pi }{4} + pi n, n in Z}$