Proving $\cos(x) = 0$ iff $x = \pi k + \pi/2$: I've hit a roadblock

Question

This is my working so far:

$$e^{ix} + e^{-ix} = 0$$

$$e^{-ix}\left(e^{2ix}+1\right) = 0$$

Therefore,

$$e^{2ix} = -1\implies e^{ix} = i$$

$$ix = \log(i)\implies ix = i\left(2\pi k + \frac\pi2\right)$$

Hence,

$$x = 2\pi k + \frac\pi2$$ (where $k$ is an integer)

I do not understand why it is $\pi k$ and not $2\pi k$

Answer 1

$$e^{2ix}=−1\implies e^{ix}=i$$

This is not correct. You could also have $e^{ix} = -i$, which is where your missing solutions are.

Answer 2

There is no need to use exponential function with complex arguments. This is a simple problem from trigonometry and usual trigonometric identities suffice. Moreover this is proved in every trigonometry textbook. You may compare the following with the treatment given in your textbook.

It is clear that if $x=\pi k +\pi/2$ then $$\cos x=\cos\pi k\cos (\pi/2)-\sin k\pi\sin(\pi/2)=(-1)^{k}\cdot 0-0\cdot 1=0$$ Let's now prove that if $x$ is not of the above form then $\cos x\neq 0$. Given any real number $x$ there is a unique integer $k$ and unique real number $y$ such that $x=\pi k+y$ and $0<y<\pi$. And if $x$ is not of the specific form desired, then $y\neq \pi/2$. Thus either $y\in(0,\pi/2)$ or $y\in(\pi/2,\pi)$. In both cases we know that $\cos y\neq 0$. Now $$\cos x=\cos \pi k\cos y-\sin\pi k\sin y=(-1)^{k}\cos y$$ Clearly the RHS is non-zero as it is a product of two non-zero numbers and hence $\cos x\neq 0$ and this completes the proof.