Cos4x=2

- If x is real: There are no solutions. cos(4x) ∈ [-1, 1], but 2 is outside this range. - If you allow complex x: Solve using e^(i4x). 1) cos(4x) = (e^{i4x} + e^{-i4x})/2 = 2 ⇒ e^{i4x} + e^{-i4x} = 4 2) Let t = e^{i4x}. Then t + 1/t = 4 ⇒ t^2 - 4t + 1 = 0 ⇒ t = 2 ± √3 3) Thus e^{i4x} = 2 ± √3. Writing ln for the complex log: i4x = ln(2 ± √3) + 2πi k, k ∈ Z 4) Therefore x = (π/2)k − (i/4) ln(2 ± √3). Since ln(2−√3) = −ln(2+√3), this can be written as x = (π/2)k ± (i/4) ln(2+√3), for any integer k. So: - Real solutions: none. - Complex solutions: x = (π/2)k ± (i/4) ln(2+√3), k ∈ Z.