Solve the following equation on the interval [0,2pi) sec^8x=-2

Question

Solve the following equation on the interval [0,2pi)

Answer 1

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 re^iθ=r(cosθ+isinθ) and r⁸e^8iθ=-½.

-½ can be expressed as ½(cosπ+isinπ)=½e^iπ.

Therefore  r⁸e^8iθ=½e^iπ, making r⁸=0.5 and θ=⅛π.

So z=0.5^⅛e^⅛iπ=cos(x).

cosh(x)=½(e^x+e^-x),

so cosh(ix)=½(e^ix+e^-ix)=½(cos(x)+isin(x)+cos(x)-isin(x))=cos(x).

cos(x)=cosh(ix)=0.5^⅛e^⅛iπ. ix=cosh^-1(0.5^⅛e^⅛iπ), x=-icosh^-1(0.5^⅛e^⅛iπ).

0.5^⅛=0.9170 approx; ⅛π=0.3927 approx. x=-icosh^-1(0.9170e^0.3927i).

cos(π/8)=0.9239, sin(π/8)=0.3827, so e^⅛iπ=0.9239+0.3827i.

x=-icosh^-1(0.8472+0.3509i).