TITLE: Solve this equation involving binomial coefficients
QUESTION [5 upvotes]: The question is that we have to find the real value(s) of $x\in[-1,1]$ satisfying the equation:
$${11 \choose 1}x^{10}-{11 \choose 3}x^{8}+{11 \choose 5}x^{6}-{11 \choose 7}x^{4}+{11 \choose 9}x^{2}-{11 \choose 11}=0$$
I couldn't spot any standard series to simplify the expression on the LHS and so I tried writing it as $$\frac{-1}{2}\big[(1+x)^{11}+(1-x)^{11}]$$ but this is obviously not true since it does provide us with the required terms but their signs are not correct.  
How can I make the appropriate signs appear on the terms too? I suspect that $i$ may be involved but I'm unsure about how to apply this idea. Any help would be appreciated.

REPLY [3 votes]: Hint:
$$\dfrac{(1+y)^{2n+1}+(1-y)^{2n+1}}2=\sum_{r=0}^{r\le n}\binom{2n+1}{2r}y^{2r}$$
We have $$\sum_{r=0}^n\binom{2n+1}{2r}(-1)^rx^{2r}=0$$
Comparing $y=ix$ 
So, $$(1+ix)^{2n+1}=-(1+ix)^{2n+1}$$
$$\iff\left(\dfrac{1+ix}{1-ix}\right)^{2n+1}=-1=e^{(2m+1)i\pi}$$