y = 2 + 2sec(2x)

The upper part of the range will be when the secant has the smallest
positive value up to infinity.

The smallest positive value of the secant is 1

So the minimum of the upper part of the range of

y = 2 + 2sec(2x) is 2 + 2(1) = 2 + 2 = 4

So the upper part of the range is [4, )

The lower part of the range will be from negative infinity
up to when the secant has the largest negative value.

The largest negative value of the secant is -1

So the maximum of the lower part of the range of

y = 2 + 2sec(2x) is 2 + 2(-1) = 2 - 2 = 0

So the lower part of the range is (, 0].

Therefore the range is (, 0] U [4, )

Edwin