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.

Answer

Assuming $x,\theta \in \mathbb{R}$, $n\in\mathbb{Z}$:

$$\sum _{k=0}^{n}{n\choose k}{x}^{k}\cos \left( k\theta \right) =\frac{1}{2} \left( 1+x{{\rm e}^{i\theta}} \right) ^{n}+\frac{1}{2} \left( 1+x{{\rm e}^{ -i\theta}} \right) ^{n}=0,$$
$$\Rightarrow\dfrac{ \left( 1+x{{\rm e}^{i\theta}} \right) ^{n}}{\left( 1+x{{\rm e}^{ -i\theta}} \right) ^{n}}=-1={\rm e}^{i\pi},$$
$$\dfrac{ \left( 1+x{{\rm e}^{i\theta}} \right)}{\left( 1+x{{\rm e}^{ -i\theta}} \right)}={\rm exp}\left({\dfrac{im\pi}{n}}\right):m \,\text{odd}\in \mathbb{Z},$$
$$x=\dfrac{\sin \left( {\dfrac {\pi m}{2n}} \right)}{\sin \left( \theta-{\dfrac {\pi m}{2n}} \right) }.$$