6cos^2 4x+cos4x-1=0 Помогите пожалуйста,буду благодарна!!!

cos4x=a 6a²+a-1=0 D=1+24=25 a1=(-1-5)/12=-1/2⇒cos4x=-1/2⇒4x=+-2π/3+2πk⇒x=+-π/6+πk/2,k∈z a2=(-1+5)/12=1/3⇒cos4x=1/3⇒4x=+-arccos1/3+2πk⇒x=+-1/4*arccos1/3+πk/2,k∈z

6cos^2 4x+cos4x-1=0 cos4x=t   !t!≤1 6t²+t-1=0 D=1+4*6=25 t12=(-1+-5)/12= -1/2  1/3 cos4x=1/3 4x=+-arccos(1/3)+2πk k⊂Z x=+-arccos(1/3)/4 + πk/2 cos4x=-1/2 4x=+-2π/3+2πk k⊂Z x=+-π/6 + πk/2