f(x)=1/(2-4cos(2x)) has a discontinuity when the denominator is zero.
Assuming you mean the interval [0,2π], 4cos(2x)=2 at the discontinuity.
So cos(2x)=½, 2x=π/3 or 60º, so x=π/6 (30º).
But cos(2x)=½ when 2x=2π-π/3=5π/3 (300º). So x can also be 5π/6 (150º). Between π/6 and 5π/6 it’s continuous.