[tex]4cos^2x +12cosx +5 = 0[/tex]

D=b²-4ac=144-80=64; √64=8;Cos(x₁)=(-b+√D)/2a=(-12+8)/8=-4/8=-1/2X₁=(2π/3)+2πk, k∈ZX₂=-(2π/3)+2πk, k∈Z;Cos (x₂)=(-b-√D)/2a=(-12-8)/8=-20/8=-5/2=-2, 5>-1⇒нет решения