3sin^2x+4cos^x-13sinx*cosx=0

3sin^2 (x) + 4cos^2 (x) - 13sin (x) · cos (x) = 0;

Разделим на cos^2 (x), cos^2 (x) ≠ 0;

3tg^2 (x) - 13tg (x) + 4 = 0;

Сделаем замену переменных:

tgx = t;

3t^2 - 13t + 4 = 0;

D = 169 - 48 = 121;

t1 = 4; ⇒ tgx = 4;

x = arctg (4) + пn, n∈Z;

t2 = 1/3; ⇒ tgx = 1/3;

x = arctg (1/3) + пk, k∈Z.

Ответ: x = arctg (4) + пn, n∈Z; x = arctg (1/3) + пk, k∈Z.