Answer:
There is no real solution for this equation. However, the solution exists in complex world.
[tex]We have,\\e^{ix} = cosx + isinx.......(1)\\e^{-ix} = cos(-x) + isin(-x)\\or, e^{-ix} = cosx - isin(x)......(2)\\Subtracting equation(2) from equation (1),\\e^{ix} - e^{-ix} = 2isin(x)\\or, sin(x) = \frac{e^{ix}-e^{-ix}}{2i}\\We need to solve:\\sin(x) = 2\\or, \frac{e^{ix}-e^{-ix}}{2i} = 2\\or, e^{ix} - e^{-ix} = 4i\\or, e^{ix} - \frac{1}{e^{ix}} = 4i\\or, e^{2ix} - 1 = 4ie^{ix}\\or, e^{2ix} - 4ie^{ix}-1=0\\\\or, (e^{ix})^{2} + (-4i) e^{ix} - 1 = 0\\[/tex]
Using quadratic formula:
