Find all angles $\theta$ such that $x_1\cos(0)+x_2\cos(\theta)+x_3\cos(2\theta)+\dots + x_{n+1}\cos (n\theta)=0$ Find all angles $\theta$ such that $\exists x_1,x_2,\dots x_{n+1}, n\in \mathbb{Z}$ such that $$x_1\cos(0)+x_2\cos(\theta)+x_3\cos(2\theta)+\dots + x_{n+1}\cos (n\theta)=0,\\x_1\sin(0)+x_2\sin(\theta)+x_3\sin(2\theta)+\dots + x_{n+1}\sin(n\theta)=0.$$
Here is my progress. If $\theta$ satisfies then $180-\theta$ satisfies too. Clearly, angles of the form $\frac{360}{r}$ satisfy ( simply consider $r$ regular polygon with external angle $\frac{360}{r}$). For example, $\cos(0)+ \cos(60)+\cos(120)+\cos(180)+\cos(240)+\cos(300)=0$
I also got $\theta= 109.47$ or $\cos(\theta)=-1/3 $ satisfying with $3cos(0)+2\cos(\theta)+3\cos(2\theta)=0.$
 A: Multiply the sine equation by the imaginary unit $i$ and then add the two equations together. Using Euler's identity $e^{i\theta}=\cos\theta+i\sin\theta$ we obtain the equivalent condition
$$x_1+x_2e^{i\theta}+x_3e^{2i\theta}+\dotsm+x_{n+1}e^{in\theta}=0.$$
But since $e^{in\theta} = (e^{i\theta})^n$, we can put $z=e^{i\theta}$ and the question becomes the following: for which complex $z$ on the unit circle are there integers $n,x_1,\dotsc, x_{n+1}$ such that
$$x_1+x_2z+x_3z^2+\dotsm+x_{n+1}z^n=0?$$
In other words, which complex $z$ on the unit circle are algebraic? So far as I know, there isn't a nice characterization of these numbers. At any rate, the set of $\theta$ which satisfy your question is given by
$$S = \bigcup_{\mathcal{A}(\mathbb{C})\cap\{|z|=1\}}{\rm arg(z)}$$
where $\mathcal{A}(\mathbb{C})$ denotes the algebraic numbers of $\mathbb{C}$ and ${\rm arg}(z) = \{{\rm Arg}(z)+2\pi k:k\in\mathbb{Z}\}$.
We can describe a (probably proper) subset of $S$ at least. If $\theta$ is a rational multiple of $\pi$, then it solves the problem. Indeed, we can write $\theta = 2\pi m/n$ for $m,n$ integers with $n$ positive and then $z=e^{i\theta}$ will be a solution of the polynomial equation $-1+z^n=0$.
