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so, cos2x=cosx, I think x is in those special angles

use identity : cos2x=2cos^2 x - 1

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..