Solve[ForAll[{x, y}, Equivalent[y == Sqrt[3]/2 - x, y + 2 (x - x0) Sin[\[Pi]/6 + 2 x0] == Cos[\[Pi]/6 + 2 x0]]], x0, Reals]
It is to find the value of x0
If just solving the equations is what you want, then this works:
eqs = {y == Sqrt[3]/2 - x, y + 2 (x - x0) Sin[\[Pi]/6 + 2 x0] == Cos[\[Pi]/6 + 2 x0]}; Solve[eqs, {x, y}]
