Problem:
What is the sum of all positive real solutions x to the equation
2cos(2x)(cos(2x)βcos(x2014Ο2β))=cos(4x)β1?
Answer Choices:
A. Ο
B. 810Ο
C. 1008Ο
D. 1080Ο
E. 1800Ο
Solution:
If x=21βΟy, then the given equation is equivalent to
2cos(Οy)(cos(Οy)βcos(y4028Οβ))=cos(2Οy)β1
Dividing both sides by 2 and using the identity 21β(1βcos(2Οy))=sin2(Οy) yields
cos2(Οy)βcos(Οy)cos(y4028Οβ)=21β(cos(2Οy)β1)=βsin2(Οy).
This is equivalent to
1=cos(Οy)cos(y4028Οβ)
Thus either cos(Οy)=cos(y4028Οβ)=1 or cos(Οy)=cos(y4028Οβ)=β1. It follows that y and y4028β 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=22β
19β
53, that is aβ{1,19,53,19β
53}. Then cos(Οy)=cos(2aΟ)= 1 and cos(y4028Οβ)=cos(a2014Οβ)=1. Therefore the sum of all solutions x is Ο(1+19+53+19β
53)=Ο(19+1)(53+1)=1080Οβ.
The problems on this page are the property of the MAA's American Mathematics Competitions