3 cos^2 x - sin^2 x + 4 sin x = 0

3cos²x-sin²x+4sinx=0

3 (1-sin²x) - sin²x+4sinx=0

3-3sin²x-sin²x+4sinx=0

-4sin²x+4sinx+3=0 |: (-1)

4sin²x-4sinx-3=0

sin x=t; |t|≤1

4t²-4t-3=0

D=16-4*4 * (-3) = 16+48=64

t1 = (4+8) / 8=12/8=1.5 (не удовл. усл. |t|≤1)

t2 = (4-8) / 8=-4/8=-1/2

sinx=-1/2

x1=-π/6+2πn; n∈z

x2=7π/6+2πn; n∈z

Ответ : x1=-π/6+2πn; n∈z

x2=7π/6+2πn; n∈z