Решить 2sin^2x+11 sin x cos x+12cos^2x=0

2Sin²x + 11SinxCosx + 12Cos²x = 0 делим на Cos²x
2tg²x + 11tgx + 12 = 0 
t = tgx ∈ [-∞;+∞]
2t² + 11t + 12 = 0
D = 121 - 4 * 2 * 12 = 25
t₁ = (-11 + 5) / 4 = -6/4 = -1.5
t₂ = (-11 - 5) / 4 = -16/4 = -4
tgx = -1.5
x = arctg(-1.5) + πn, n∈Z
tgx = -4
x = arctg(-4) + πn, n∈Z