Question 391582
{{{cos 2x = cos x}}} <==> {{{2(cos x)^2 - 1 = cosx}}} <==> {{{2(cos x)^2 - cosx - 1 = 0}}} <==> (2 cosx + 1)(cosx - 1) = 0 ==> cosx = -1/2 or cos x = 1.
From cosx = -1/2, x = {{{-2pi/3}}},{{{ 2pi/3}}}.
From  cos x = 1, x = 0. 
The solutions are then 0,{{{-2pi/3}}},{{{ 2pi/3}}}.