Question

8sin²x + 2sinxcosx - 6cos²x = 0

Answer 1

Разделив на cos^2x, получим
8tg^2x + 2tgx - 6 = 0 |:2
4tg^2x + tgx - 3 = 0
D=b^2-4ac = 1 + 4 * 4 * 3 = 49
tgx = (-1+7)/8 = 3/4
x1 = arctg(3/4) + pi*n

tgx = (-1-7)/8 = -1
x2 = -pi/4 + pin