Posted by Abbey on Sunday, February 5, 2012 at 11:08pm.
Find all solutions to the equation in the interval [0,2pi)
Cosxcos2x

calculus  Reiny, Monday, February 6, 2012 at 12:45am
I will assume your "equation" is
cosx  cos 2x = 0
cosx  (2cos^2 x  1) = 0
2cos^2 x  cosx  1 = 0
(2cosx + 1)(cosx  1) = 0
cosx = 1/2 or cosx = 1
if cosx = 1/2 , then
x = 2π/3 or x = 4π/3 , (120° or 240°)
if cosx = 1 , then
x = 0 or 2π , (0° or 360°)

calculus  Abbey, Monday, February 6, 2012 at 8:12am
Thanks!
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