Problem:
Find the sum of the values of x such that cos33x+cos35x=8cos34xcos3x, where x is measured in degrees and 100<x<200.
Solution:
The given equation implies that
cos33x+cos35x=(2cos4xcosx)3=(cos(4x+x)+cos(4x−x))3=(cos5x+cos3x)3.
Let y=cos3x and z=cos5x. Then y3+z3=(y+z)3. Expand and simplify to obtain 0=3yz(y+z). Thus y=0 or z=0 or y+z=0, that is, cos3x=0 or cos5x=0 or cos5x+cos3x=0. The solutions to the first equation are of the form x=30+60j, where j is an integer; the second equation has solutions of the form x=18+36k, where k is an integer. The third equation is equivalent to cos4xcosx=0, so its solutions are of the form x=2221+45m and x=90+180n, where m and n are integers. The solutions in the interval 100<x<200 are 150,126,162,198,11221, and 15721, and their sum is 906.
The problems on this page are the property of the MAA's American Mathematics Competitions