Question

$4cos^2x +12cosx +5 = 0$

Answer 1

D=b²-4ac=144-80=64; √64=8;
Cos(x₁)=(-b+√D)/2a=(-12+8)/8=-4/8=-1/2
X₁=(2π/3)+2πk, k∈Z
X₂=-(2π/3)+2πk, k∈Z;
Cos (x₂)=(-b-√D)/2a=(-12-8)/8=-20/8=-5/2=-2, 5>-1⇒нет решения