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Why is it the most energy efficient to change orbit inclination while crossing the equator?
Specifically, it's most efficient to do a plane change at one of the two "nodes" where the origin orbital plane intersects the destination plane. ANASIS-II is destined for geostationary orbit, so its destination plane is the plane of the equator.
Any orbit around a single massive body lies in a single plane. It should be clear that you can't enter an equatorial-plane orbit at any point except a point directly over the equator. Coming from any non-equatorial orbit, there are two points on the orbit where the planes intersect. If you try and do a burn to enter a particular destination orbit from anywhere else, you just push the intersection point a little further around the orbit.
(It's super easy to demonstrate this in Kerbal Space Program, but kind of hard to put into words!) //
A great aid to intuition is to remember one principle about orbit changes: if the engine is off, the orbiter always returns to same point one orbit later.
So for any orbit change, if you want to do only a short burn, it has to be at a point that is common for both the current orbit and the destination orbit. This applies to inclination changes, altitude changes and basically any orbit change. If the orbits do not have a common point, the change requires two burns and an intermediate orbit, such as Hohmann transfer orbit.
Like Russell's answer details, for a geostationary target orbit those common points are always above equator. For e.g. polar target orbit, the points would be somewhere else.