Find two roots of unity such that their sum equal 1 Okay so what I did, is that I said that: $$\exp((2i\pi)/n)+\exp((2i\pi)/n)=1$$ which was equivalent to $$2\exp((2i\pi)/n)=1$$
$$\exp((2i\pi)/n)=\frac 1 2$$ But then only non-sense came from this I don't know why.
If $w$ is a root of unity then $w^n=1\implies |w|=1$. So $w$ belongs to the unit circle of radius $1$ centred in $(0,0)$.
Let assume $1-w$ is also a root of unity, then $|1-w|=1$ and $w$ also belongs to the circle of radius $1$ centred in $(1,0)$.
The intersection of these two circles have abscissa $\dfrac 12$ and correspond to the angle $\dfrac{\pi}3$.
[since $\cos(\frac\pi3)=\frac 12$].
And you can verify that $w=e^{i\pi/3}=\dfrac{1+i\sqrt{3}}2$ and $1-w=e^{-i\pi/3}=\dfrac{1-i\sqrt{3}}2$
both verify $w^6=1$ and have sum $1$.
Regarding your approach you should solve instead $e^{i2k_1\pi/n}+e^{i2k_2\pi/n}=1$.
Here you assumed wrongly $k_1=k_2=1$.
By developping this we get $\begin{cases}\cos(\frac{2k_1\pi}n)+\cos(\frac{2k_2\pi}n)=1\\\sin(\frac{2k_1\pi}n)+\sin(\frac{2k_2\pi}n)=0\end{cases}$
The second equation forces $\theta_1=\theta_2+\pi$ or $\theta_1=-\theta_2$
In the first case, the cosinus sum also vanish, thus we have $2\cos(\frac{2k_1\pi}n)=1\iff \dfrac{2k_1\pi}n=\pm\dfrac{\pi}3$
And we get the same solution than previously.
Having derived $2\Re[\exp(i2\pi/n)]=1$, proceed as follows, starting with Euler's Formula:
$2\cos(2\pi/n)=1$
$2\cos(2\pi/n)\sin(2\pi/n)=\sin(2\pi/n)$
$\sin[2(2\pi/n)]=\sin(\pi/2n)$ (from the double-angle formula for sines or the sum-product relation for $\cos u\sin v$)
So $2(2\pi/n)-2\pi/n=2\pi/n$ is an even number times $\pi$ or else $2(2\pi/n)+2\pi/n=6\pi/n$ is an odd number times $\pi$. The only whole numbers for $n$ which satisfy at least one of these relations are $1,2,6$, and just one of those three candidates actually survives checking.
