2014 AMC 12 B problem 25 What is the sum of all positive real solutions $x$ to the following equation? $$2\cos(2x)\left( \cos(2x) - \cos{\left(\frac{2014\pi^2}{x^2}\right)} \right) = \cos(4x) - 1 $$
 A: Using $\cos2A=2\cos^2A-1,$
$$2\cos^22x-2\cos2x\cos\left(\frac{2014\pi^2}{x^2}\right)=2\cos^22x-1-1$$
$$\implies\cos2x\cos\left(\frac{2014\pi^2}{x^2}\right)=1$$
As for real $y,-1\le\cos y\le1;$ We need 
$$\cos2x=\cos\left(\frac{2014\pi^2}{x^2}\right)=\pm1$$
Taking the '+' sign, 
$\displaystyle\cos2x=1\implies 2x=2n\pi\iff x=n\pi$ where $n$ is an integer
and $\displaystyle\cos\left(\frac{2014\pi^2}{x^2}\right)=1\implies \frac{2014\pi^2}{x^2}=2m\pi\iff x^2=\frac{1007\pi}m$  where $m$ is an integer
$\displaystyle\implies \frac{1007\pi}m=(n\pi)^2\iff\pi=\frac{1000}{mn^2}$ which is rational unlike $\pi$
Can you manage the '-' sign from here?
A: After a straightforward application of the formula $\cos{2A = 2\cos^2{A} - 1}$ to the right side and some algebraic simplification, we obtain 
$$\cos{(2x)}\cos{\left( \frac{2014\pi^2}{x^2} \right)}  = 1$$
Note that we cannot have $\cos 2x \in (0, 1)$ for that would mean that the other factor was greater than 1, clearly impossible. Similarly, we cannot have $\cos{2x} \in (-1, 0)$. Thus, $$\cos(2x) = \cos{\left( \frac{2014\pi^2}{x^2}\right)} = \pm1  $$
The only viable case is where $2014\pi/k$ is a whole multiple of $\pi$. Thus, the solutions are $\pi, 19\pi, 53\pi, 1007\pi$, which add up to $1080\pi$.
A: It seems to not be a hard question!
$2\cos(2x)\left( \cos(2x) - \cos{\left(\frac{2014\pi^2}{x^2}\right)} \right) = \cos(4x) - 1=2cos^{2}(2x)-2 $$\Longrightarrow$ $\cos(2x)\cos(\frac{2014\pi^2}{x^2})=1$
the hint is :
$\cos(\frac{2014\pi^2}{x^2})=sec(2x)$
and $cos\in[0,1]$ and $sec\in[1,\infty]$
Is it helpful?
