sec8(x)=-2 can have no real solution because sec8(x)=(sec4(x))2 and all squares are positive.
If we assume a complex solution rather than error in the question:
Let z=cos(x), then z8=-½.
z can also be expressed as reiθ=r(cosθ+isinθ) and r8e8iθ=-½.
-½ can be expressed as ½(cosπ+isinπ)=½eiπ.
Therefore r8e8iθ=½eiπ, making r8=0.5 and θ=⅛π.
So z=0.5⅛e⅛iπ=cos(x).
cosh(x)=½(ex+e-x),
so cosh(ix)=½(eix+e-ix)=½(cos(x)+isin(x)+cos(x)-isin(x))=cos(x).
cos(x)=cosh(ix)=0.5⅛e⅛iπ. ix=cosh-1(0.5⅛e⅛iπ), x=-icosh-1(0.5⅛e⅛iπ).
0.5⅛=0.9170 approx; ⅛π=0.3927 approx. x=-icosh-1(0.9170e0.3927i).
cos(π/8)=0.9239, sin(π/8)=0.3827, so e⅛iπ=0.9239+0.3827i.
x=-icosh-1(0.8472+0.3509i).