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

cos4x=a6a²+a-1=0D=1+24=25a1=(-1-5)/12=-1/2⇒cos4x=-1/2⇒4x=+-2π/3+2πk⇒x=+-π/6+πk/2,k∈za2=(-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=0cos4x=t   !t!≤16t²+t-1=0D=1+4*6=25t12=(-1+-5)/12= -1/2  1/3cos4x=1/34x=+-arccos(1/3)+2πk k⊂Zx=+-arccos(1/3)/4 + πk/2cos4x=-1/24x=+-2π/3+2πk k⊂Zx=+-π/6 + πk/2