Solve for x in the set [0;2pi). cos2x=1/2

cos 2x = 1/2 <=> 2x = arccos
(1/2)

2x = pi/3

x =
pi/6

The first value of x is located in the 1st
quadrant.

Since we have to determine all convenient values
from all 4 quadrants, we'll step in the 2nd quadrant.

x =
pi - pi/6

x = 5pi/6

We'll
determine the equivalent value in the 3rd quadrant.

x = pi
+ pi/6

x = 7pi/6

We'll
determine the equivalent value in the 4th quadrant.

x = 2pi
- pi/6

x =
11pi/6

The all 4 possible values of x are:
{pi/6 ; 5pi/6 ; 7pi/6 ; 11pi/6}.