solve the inequality for 0≤ x≤ 2pi and express your answer in interval notation cos(2x)>1/2?

0 ≤ x ≤ 2π
Then, 0 ≤ 2x ≤ 4π

Recall that :
1. Cosine is positive in the first and fourth quadrants.
2. cos(0) = 1 and cos(π/3) = 1/2

cos(2x) > 1/2
0 ≤ 2x < π/3 or [2π - (π/3)] < 2x < [2π + (π/3)] or [4π - (π/3)] < 2x ≤ 4π
0 ≤ 2x < π/3 or 5π/3 < 2x < 7π/3 or 11π/3 < 2x ≤ 4π
0 ≤ x < π/6 or 5π/6 < x < 7π/6 or 11π/6 < x ≤ 4π

Hence, x ∈ [0, π/6) ∪ (5π/6, 7π/6) ∪ (11π/6, 2π]

i) Using multiple angle identity, the given one is: 2*cos²x - 1 > 1/2

ii) ==> cos²x > 3/4
So |cos(x)| > √3/2

Hence cos(x) > √3/2 or cos(x) < -√3/2

iii) Considering cos(x) > √3/2, cos(x) is in (√3/2, 1]
==> x is in [0, π/6) U (11π/6, 2π]

Considering cos(x) < -√3/2, cos(x) is in [-1, -√3/2)
==> x is in (5π/6, 7π/6)

Combining both x is in [0, π/6) U (5π/6, 7π/6) U (11π/6, 2π]

[0 , π / 6 ] U [ 11 π / 6 , 2π ]