1) 3sinx*cosx+4cos^2x=0 2) 3sin^2x+4cos^2x=13cosx*sinx

1)

3sinx * cosx + 4cos^2x = 0; cosx (3sinx + 4cosx) = 0; [cosx = 0;

[3sinx + 4cosx = 0; [cosx = 0;

[tgx = - 4/3; [x = π/2 + πk, k ∈ Z;

[x = - arctg (4/3) + πk, k ∈ Z.

2)

3sin^2x + 4cos^2x = 13cosx * sinx; 3sin^2x - 13cosx * sinx + 4cos^2x = 0; 3tg^2x - 13tgx + 4 = 0; D = 13^2 - 4 * 3 * 4 = 169 - 48 = 121 = 11^2; tgx = (13 ± 11) / 6;

a) tgx = (13 - 11) / 6 = 2/6 = 1/3;

x = arctg (1/3) + πk, k ∈ Z.

b) tgx = (13 + 11) / 6 = 24/6 = 4;

x = arctg (4) + πk, k ∈ Z.

Ответ:

1) π/2 + πk; - arctg (4/3) + πk, k ∈ Z; 2) arctg (1/3) + πk; arctg (4) + πk, k ∈ Z.