Question

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

Answer 1

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

Answer 2

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