Not without knowing anything about the map projection used; it could require something hideously complicated or it might only need a bit of linear scaling.
Assuming that the sides of the image are parallel to the lines of latitude and longitude, I suppose linear scaling would be close enough for something that small ... you'll need to know the latitudes and longitudes of the corners of your image, remembering that in the northern hemisphere the latitude increases the further north you go, while images number rows of pixels going down. Hm. I can never remember this off the top of my head, I have to dig back to my memories of school and rederive it from first principles. I'm just going to do the longitude.
Lessee - leftmost edge has x=0, lon=L1°; rightmost edge has x=X, lon=L2°.
Any pixel in beween has x between 0 and X, and lon between L1° and L2°.
Scaling both to be between 0 and 1 means dividing x by X; it also means subtracting L1 from lon and then
dividing by (L2-L1).
(Okay, if I wanted to be picky, it means subtracting 0 from x and then dividing by (X-0). That's just to show the same thing is being done to x and lon - it's just that with x most of the bits fall out 'cos they're zero.)
If they're both scaled to the same range they should end up the same (because they're both referring to the same point).
x/X = (lon-L1)/(L2-L1)
We've got x, and we want lon, so we solve for that:
x/X = (lon-L1)/(L2-L1)
(L2-L1)*x/X = lon-L1
(L2-L1)*x/X+L1 = lon
There you go, to convert from x to lon it's
lon = (L2-L1)*x/X+L1