Another approach. And[] is not equivalent to Equivalent[], of course, but I wasn't sure that mattered.
Eliminate[
{y == Sqrt[3]/2 - x,
y + 2 (x - x0) Sin[\[Pi]/6 + 2 x0] == Cos[\[Pi]/6 + 2 x0]},
y]
SolveAlways[%, {x}]
Solve[Flatten[% /. Rule -> Equal], x0, Reals]
(*
Sqrt[3] - 2 x - 2 Cos[\[Pi]/6 + 2 x0] + 4 x Sin[\[Pi]/6 + 2 x0] - 4 x0 Sin[\[Pi]/6 + 2 x0] == 0
{{Cos[\[Pi]/6 + 2 x0] -> 1/2 (Sqrt[3] - 2 x0), Sin[\[Pi]/6 + 2 x0] -> 1/2}}
{{x0 -> 0}}
*)
You ask "why?" It's often the case that the real desire is for a workaround. And in this case, why may be hard to pin down precisely. Here's a stab at a rough explanation. Suppose
$T(x)$ is some transcendental function and
$F(u,v)$ is somewhat simpler, say, a polynomial or rational function. Then often equations of the form
$F(x, T(x))=0$ are difficult to solve. For instance, if
$T$ is the exponential function, then some equations can be solved in terms of the Lambert
$W$ function, a.k.a. ProductLog[]. Sometimes functions have been invented to solve such equations, especially when there's a nice theory or algorithm to support their use. Sometimes such functions or algorithms are yet to be invented. For an equation like yours, which significantly has x0 both inside and outside transcendental functions, whatever algorithms are used, none hit upon a solution. It seems odd to humans when there is a fairly easy human way to see how to solve the particular case at hand.
