A faster solution to this equation: $e^{-ix}(1+e^{-ix}) = -1$

Question

I want to find $x$ such that

$$e^{-ix}(1+e^{-ix}) = -1$$

I was able to see that it was the same as $\cos x = -0.5$ with a bit of work, but I think there must be a faster way since usually questions from GATE exam have elegant solutions.

Let me give some context, I needed to solve this equation in order to find the frequencies rejected by a digital system. Question 10.2 (GATE IN 2003 digital systems rejection of frequencies cause gain is zero)

The following is my attempt

So I wanted an alternative way to solve it, it would be nice if it was faster by hand and also any fast way to find the general form for x once we it's a set of angles?

Answer 1

This just a quadratic equation in $z=e^{-ix}$, where $\cos(x)=\operatorname{Re}(z)$. Use the quadratic formula to solve $$z^2+z+1=0,$$ to find that $z=\frac{-1\pm\sqrt{-3}}{2}=-\frac{1}{2}\pm\frac{1}{2}\sqrt{3}i$, and hence that $\cos(x)=\operatorname{Re}(z)=-\frac{1}{2}$.

Answer 2

Just take modulus and square the equation. You will get $\cos\, x=-0.5$ immediately.

$$|e^{-ix}|^{2} |1+e^{-ix}|^{2}= |-1|^{2}$$

$$1 \times \left ( (1+\cos x)^2 +(-\sin x)^2 \right ) = 1$$

$$ 2 + 2 \cos x = 1 $$

So $\cos\, x=\frac{1-2}{2}$!!

Answer 3

$$\begin{align} e^{-ix}\left(1+e^{-ix}\right)=-1 &\quad\to\quad1 + e^{-ix} = -e^{ix} \\[6pt] &\quad\to\quad e^{ix}+e^{-ix}=-1 \\[6pt] &\quad\to\quad \frac12\left(e^{ix}+e^{-ix}\right) = -\frac12 \\[6pt] &\quad\to\quad \cos x = -\frac12 \end{align}$$