3sin²x+4cos²x-13sinx∗cosx=0 / (cos^2x)
3tg^2x - 13tgx + 4 = 0
D = 169 - 4*3*4 = 121
1) tgx = (13 - 11)/6
tgx = 1/3
x1 = arctg(1/3) + πn, n∈Z
2) tgx = (13 + 11)/6
tgx = 4
x2 = arctg(4) + πk, k∈Z
Ответ: x1 = arctg(1/3) + πn, n∈Z; x2 = arctg(4) + πk, k∈Z.