The problem here is given a point on a sphere map it to a 2x1 texture. The texture is an equirectangular since the x and y values linearly correlate to the longitude and latitude.
First normalize the position on the sphere so it is a unit sphere which saves some math later on. To find the latitude in radians it is based off arcsin of the y value divided by the length of the vector, which is 1 since it is normalised, so the value is arsin(y/1) which is just arcsin(y). This technically gives the value between -PI/2 and PI/2. Dividing by PI scales it to -0.5 to 0.5, so adding it from 0.5 scales it between 0 and 1. The only prolem is that 0.5+(asin(y)/PI) is upside down since in GLSL the texture sampler uses 0,0 as the top left corner not bottom right, so to flip it do 1-0.5-(asin(y)/PI), which just becomes 0.5-(asin(y)/PI).
The longitude is more complicated. Finding the angle around the y axis can be accomplished by the built in atan2 function which automatically finds the correct quatrant since x and z are passed in seperately, so it can look at their signs. Atan2(x, z) will give the angle around between -PI and PI covering the whole circle instead of 0 to 2PI, but this doesn't really matter because OpenGl wrapps the texutres around so if the x or y value go off the range of 0-1 it just repeats an infinite grid. This works for longitude but not latitude since for latitude it must also be flipped since the north pole doesn't come out the south pole when crossing. Going ahead and dividing by 2PI scales it, but the texture I used is flipped so I make it negative. The end equation is -(atan2(z, x)/(2PI)).
vec3
globe(vec3 mpos)
{
float cf = 1.0;
mpos = normalize(mpos);
float x = mpos.x;
float y = mpos.y;
float z = mpos.z;
float a = -(atan2(z, x) / (2.0 * PI));
float b = 0.5 - (asin(y) / PI);
return texture(sampler, vec2(a, b)).rgb * cf;
}