f (x)= x - 2sinx, 0<x<2pi (less than or equal to)

Regarding this problem, I think that I have it solved. I just need someone to check my work to make sure I got the correct answer. Here is my answer:

f ' (x)= 1 - 2cosx ; 1-2cosx=0; cosx= 1/2

cosx > 1/2

f is increasing from (0, pi/3) and (5pi/3, 2pi)

f is decreasing from (pi/3, 5pi/3)

f ''(x) = 2sinx; sinx = 0; sinx>0

(0 , 2pi)

concave up: (pi, 2pi)

concave down: (0, pi)

Feedback would be appreciated, thanks.