AMC12 2014 B

Answer (D): If $x=\frac{1}{2}\pi y$, then the given equation is equivalent to $$2\cos(\pi y)\left(\cos(\pi y)-\cos\left(\frac{4028\pi}{y}\right)\right)=\cos(2\pi y)-1.$$ Dividing both sides by $2$ and using the identity $\frac{1}{2}(1-\cos(2\pi y))=\sin^2(\pi y)$ yields $$\cos^2(\pi y)-\cos(\pi y)\cos\left(\frac{4028\pi}{y}\right)=\frac{1}{2}(\cos(2\pi y)-1)=-\sin^2(\pi y).$$ This is equivalent to $$1=\cos(\pi y)\cos\left(\frac{4028\pi}{y}\right).$$ Thus either $\cos(\pi y)=\cos\left(\frac{4028\pi}{y}\right)=1$ or $\cos(\pi y)=\cos\left(\frac{4028\pi}{y}\right)=-1$. It follows that $y$ and $\frac{4028}{y}$ are both integers having the same parity. Therefore $y$ cannot be odd or a multiple of $4$. Finally, let $y=2a$ with $a$ a positive odd divisor of $4028=2^2\cdot 19\cdot 53$, that is $a\in\{1,19,53,19\cdot 53\}$. Then $\cos(\pi y)=\cos(2a\pi)=1$ and $\cos\left(\frac{4028\pi}{y}\right)=\cos\left(\frac{2014\pi}{a}\right)=1$. Therefore the sum of all solutions $x$ is $$\pi(1+19+53+19\cdot 53)=\pi(19+1)(53+1)=1080\pi.$$