pre calc
posted by caitlyn .
cos2x=cosx in interval of [0,2pie]

cos2xcosx=0
expand the left hand side
cos²(x)sin²(x)cos(x)=0
2cos²(x)1  cos(x)=0
substitute c=cos(x)
2c²c1=0
c=(1±√(9))/4
=1 or 1/2
cos(x)=1 when x=π (0≤x≤2π)
or
cos(x)=1/2 when x=π/3 or x=5π/3 (0≤x≤2π)
Substitute each of the three solution into the original equation to make sure that the solutions are acceptable. 
Reiny is right.
There was a mistake in the solution of the quadratic.
c=(1±√(9))/4
=1 or 1/2
cos(x)=1 when x=0 or 2π (0≤x≤2π)
or
cos(x)=1/2 when x=π±π/3 (0≤x≤2π)
Substitute each of the three solution into the original equation to make sure that the solutions are acceptable.