SOLUTION: Solve cos2x = -cosx over the interval 0 ≤ x < 2pi

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Question 1138018: Solve cos2x = -cosx over the interval 0 ≤ x < 2pi
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
cos%282x%29+=+-cos%28x%29 for the interval 0+%3C=+x+%3C+2pi
cos+%282x%29%2Bcos+%28x%29+=0..........use the following identity : cos++%282x+%29=+-1%2B2+cos%5E2+%28x%29
%28-1%2B2cos%5E2+%28x%29%29%2Bcos+%28x%29+=0
2cos%5E2+%28x%29+%2Bcos+%28x%29-1+=0.......let cos+%28x%29+=u
2u%5E2++%2Bu-1+=0
2u%5E2+-u+%2B2u-1+=0
%282u%5E2+-u+%29%2B%282u-1%29+=0
u%282u+-1+%29%2B%282u-1%29+=0
%28u%2B1%29%282u-1%29+=0

solutions:
if %28u%2B1%29=0=>u=-1
if+%282u-1%29+=0=> u=1%2F2

since cos+%28x%29+=u, we have
cos+%28x%29+=-1 or
cos+%28x%29+=1%2F2

if cos+%28x+%29=-1, and 0%3C=x%3C2pi, then x=pi+
if cos+%28x+%29=1%2F2, and+0%3C=x%3C2pi, than x=pi%2F3, x=5pi%2F3+

combine all solutions: x=pi%2F3, x=5pi%2F3+,x=pi+