Problem:
What is the sum of all positive real solutions x to the equation
2cos(2x)(cos(2x)−cos(x2014π2))=cos(4x)−1?
Answer Choices:
A. π
B. 810π
C. 1008π
D. 1080π
E. 1800π
Solution:
If x=21πy, then the given equation is equivalent to
2cos(πy)(cos(πy)−cos(y4028π))=cos(2πy)−1
Dividing both sides by 2 and using the identity 21(1−cos(2πy))=sin2(πy) yields
cos2(πy)−cos(πy)cos(y4028π)=21(cos(2πy)−1)=−sin2(πy).
This is equivalent to
1=cos(πy)cos(y4028π)
Thus either cos(πy)=cos(y4028π)=1 or cos(πy)=cos(y4028π)=−1. It follows that y and y4028 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=22⋅19⋅53, that is a∈{1,19,53,19⋅53}. Then cos(πy)=cos(2aπ)= 1 and cos(y4028π)=cos(a2014π)=1. Therefore the sum of all solutions x is π(1+19+53+19⋅53)=π(19+1)(53+1)=1080π.
The problems on this page are the property of the MAA's American Mathematics Competitions