We'll transform the sum into a product using the formula:

`cos a + cos b = 2cos[(a+b)/2]*cos[(a-b)/2]`

Let `a = x + 2pi and b = x - 2pi`

`a + b = x + 2pi + x - 2pi = 2x`

`a - b = x + 2pi - x + 2pi`

`a - b = 4pi`

`cos (x + 2pi) + cos (x - 2pi) = 2 cos (2x/2)*cos (4pi/2)`

`cos (x + 2pi) + cos (x - 2pi) = 2 cos x*cos 2pi`

`But cos 2pi = 1`

`cos (x + 2pi) + cos (x - 2pi) = 2 cos x`

The sum can be calculated using the following identities:

`cos (x + 2pi) = cos x*cos 2pi - sin x*sin 2pi`

Since `sin 2pi = 0` , the product `sinx*sin 2pi` will be 0.

`cos (x + 2pi) = cos x`

`cos (x - 2pi) = cos x*cos 2pi + sin x*sin 2pi`

`cos (x - 2pi) = cos x`

`cos (x + 2pi) + cos (x - 2pi) = cos x + cos x = 2 cos x`

**Therefore, the requested sum is `cos (x + 2pi) + cos (x - 2pi) = 2 cos x.` **

cos(x+2π) + cos(x-2π)

= cos(2π+x) + cos-(2π-x)

= cos (2π+x) + cos(2π-x)

= cosx + cosx

= 2cosx