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

3(1-sin^2x)-sin^2x+4sinx=0,

3-4sin^2x+4sinx=0,

4sin^2x-4sinx-3=0,

sinx=t,

4t^2-4t-3=0,

D=64,

t1=-1/2,

t2=1.5>1

sinx=-1/2,

x=(-1)^k arcsin(-1/2)+pi*k, kєZ,

x=(-1)^(k+1) arcsin(1/2)+pi*k, kєZ,

x=(-1)^(k+1) pi/6+pi*k, kєZ

Verified answer

3cos^2x-sin^2x+4sinx=0 

4sinx^2-4sinx-3=0

 4x2 - 4x - 3 = 0
D = b2 - 4ac
D = 16 + 48 = 64 = 8^2

x1,2 = -b ± √D/2a
x1 = 4 + 8/8 = 12/8 = 3/2
x2 = 4 - 8/8 = - 4/8 = - 1/2
Ответ:  x2 = - 1/2

sin(x2)=-0,5