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Find all solutions to the equation in the interval [0, 2π). cos 2x - cos x = 0

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so, cos2x=cosx, I think x is in those special angles
use identity : cos2x=2cos^2 x - 1
so, we have a quadratic equation in form cosx, 2cos^2 x - 1 - cosx = 0 or 2cos^2 x - cosx - 1 = 0 can u factor out it

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Other answers:

ok I get cos x(2 cos x-1)=1
np, it should be (2cosx + 1)(cosx - 1) = 0
now, for zeroes take each factors be zero, so 2cosx+1=0 or cosx-1=0 solve for x
ok I got the solutions 0, 2pi/3 and 4pi/3
thank you
but, i miss one root :)
looks 0 be same for 2pi, right ?
I got the answer right
ok, great..

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