Find the sum of all the positive solutions of\[2\cos(2x) \left(\cos(2x) - \cos\left( \frac{2014\pi^2}{x} \right)\right) = \cos(4x) - 1\] $\textbf{(A)}\ \pi \qquad$ $\textbf{(B)}\ 810\pi \qquad$ $\textbf{(C)}\ 1008\pi \qquad$ $\textbf{(D)}\ 1080 \pi \qquad$ $\textbf{(E)}\ 1800\pi$
Solution
Using $\cos4x=2\cos^22x-1$, the equation becomes $2\cos^22x-2\cos2x\cos\left(\frac{2014\pi^2}x\right)=2\cos^22x-2$, i.e. $\cos2x\cos\left(\frac{2014\pi^2}x\right)=1$.
This forces both factors to equal $1$ or both to equal $-1$. The latter needs $(2n+1)(2m+1)=\frac{2\cdot2014\pi^2}{\pi^2}=4028$, impossible since a product of two odd numbers is odd.
The former gives $x=m\pi$ with $mn=1007=19\times53$ for positive integers $m,n$; divisors of $1007$ are $1,19,53,1007$, summing to $1080$.
