SOLUTION: How do you solve for x in cos2x = cosx, -pi<=x<=pi?

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Question 391582: How do you solve for x in cos2x = cosx, -pi<=x<=pi?
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
cos+2x+=+cos+x <==> 2%28cos+x%29%5E2+-+1+=+cosx <==> 2%28cos+x%29%5E2+-+cosx+-+1+=+0 <==> (2 cosx + 1)(cosx - 1) = 0 ==> cosx = -1/2 or cos x = 1.
From cosx = -1/2, x = -2pi%2F3,+2pi%2F3.
From cos x = 1, x = 0.
The solutions are then 0,-2pi%2F3,+2pi%2F3.