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