Solve for x in the set [0;2pi).

cos2x=1/2

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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}.**

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