Solving an equation involving binomial coefficients and complex numbers Question:
Solve the following equation for $x$:
$$\sum_{k=0}^{n}\binom{n}{k}x^{k}\cos(k\theta )=0$$
Attempt:
I think this equation come from:
$$(x\cos\theta+ix\sin\theta)^{k}$$
Is that right?
I don't know what to do after that.
 A: Assuming $x,\theta \in \mathbb{R}$, $n\in\mathbb{Z}$:
$$\sum _{k=0}^{n}{n\choose k}{x}^{k}\cos \left( k\theta \right) =\frac{1}{2}
 \left( 1+x{{\rm e}^{i\theta}} \right) ^{n}+\frac{1}{2} \left( 1+x{{\rm e}^{
-i\theta}} \right) ^{n}=0,$$
$$\Rightarrow\dfrac{
 \left( 1+x{{\rm e}^{i\theta}} \right) ^{n}}{\left( 1+x{{\rm e}^{
-i\theta}} \right) ^{n}}=-1={\rm e}^{i\pi},$$
$$\dfrac{
 \left( 1+x{{\rm e}^{i\theta}} \right)}{\left( 1+x{{\rm e}^{
-i\theta}} \right)}={\rm exp}\left({\dfrac{im\pi}{n}}\right):m \,\text{odd}\in \mathbb{Z},$$
$$x=\dfrac{\sin \left( {\dfrac {\pi m}{2n}} \right)}{\sin \left( 
\theta-{\dfrac {\pi m}{2n}} \right) }.$$
A: $$\sum_{0\le r\le n}\binom nkx^ke^{ik\theta}=(1+xe^{i\theta})^n=(1+x\cos\theta+ix\sin\theta)^n$$
Let $1+x\cos\theta=R\cos\alpha,  x\sin\theta=R\sin\alpha$
$$R^2=(1+x\cos\theta)^2+(x\sin\theta)^2=1+x^2+2x\cos\theta$$
$$\implies \cos\alpha=\frac{1+x\cos\theta}{\sqrt{1+x^2+2x\cos\theta}}$$
$$\implies (1+x\cos\theta+ix\sin\theta)^n=\{R(\cos\alpha+i\sin\alpha)\}^n=R^n(\cos n\alpha+i\sin n\alpha)$$
So, we need either $R=0$ or $\cos n\alpha=0$ 
If $R=0, x\sin\theta=1+x\cos\theta=0$
Case $1:$ If $x=0, 1+x\cos\theta=1\ne0\implies x\ne0$
Case $2:$ If $\sin\theta=0,\cos\theta=\pm1,1\pm x=0\implies x=\mp1 $
Else $R\ne0 ,\cos n\alpha=0\implies n\alpha=(2m+1)\frac{\pi}2$ where $m$ is any integer
Now we need to solve for $x,\theta$ from $$\cos (2m+1)\frac{\pi}{2n}=\frac{1+x\cos\theta}{\sqrt{1+x^2+2x\cos\theta}}$$
