How does $e^{5ix} = -1 $ have 5 solutions?

Question

Part of the solution to a question in my book says $e^{5ix} = -1 $ has five solutions for x. There is no further explanation. How do I arrive at this result?

Answer 1

$$e^{i5x} = -1$$ is equivalent to $$\cos(5x) + i\sin(5x) = -1$$ because of Euler's formula. This obviously has infinite solutions for $n\geq1$ along $x=\pi\frac{2n-1}{5}$. I think it says $5$ because it has $5$ solutions for $0\leq x \lt 2\pi$.

Answer 2

If $e^{5ix}=e^{iy}$

$5ix=iy+2n\pi i$ where $n$ is any integer

$x=\dfrac{y+2n\pi}5,n=0,1,2,3,4$

Here one of the possible values of $y$ is $\pi$

Answer 3

$e^{i\pi} = e^{3i\pi} = ... = e^{(2n-1)i\pi} = -1$

This has many solutions for,$$ 5x = 2n-1$$ $$x = \frac{2n - 1}{5}, n\epsilon Z$$ e.g.,

$ x = \frac{1}{5}$ or $ x = \frac{3}{5}$ or $ x = \frac{5}{5}$ or $ x = \frac{7}{5}$ or $ x = \frac{9}{5}$

Answer 4

One solution is $$5x=\pi \implies x=\pi /5$$

Other solution come from $$ 5x= 3\pi, 5\pi , 7\pi , 9\pi\implies x=3\pi/5, 5\pi/5, 7\pi/5, 9\pi/5$$