The following transforms the equation into a polynomial system, so theoretically, Solve[] will think it's solvable. However, I don't know if it's practically solvable. For me it's not, since I'm unwilling to let it run long enough to find out.
I post it in case you want to play with it. I save a denominator ghD in case it's needed to rule out spurious solutions should any solutions ever be found. The system could be simplified with assumptions (e.g., U or V could be eliminated by assuming positivity, which would reduce the degree of the system).
Note r is given by ArcTan[X, Y] if solutions for X and Y are ever found.
gh = Simplify[TrigExpand[g + h] /. {Cos[r] -> X, Sin[r] -> Y},
X^2 + Y^2 == 1]
{ghN, ghD} = NumeratorDenominator[gh]
xf = Transpose@Block[{i = 0},
DeleteDuplicates@Cases[ghN, Power[_, 1/2], Infinity] /.
Power[u_,
1/2] :> {Power[u, 1/2] -> {U, V}[[++i]], {U, V}[[i]]^2 == u}
]
sys = Join[Last[xf], {ghN == 0 /. First[xf], X^2 + Y^2 == 0}]
(*
{U^2 == (x - a X + c Y)^2, V^2 == (-d + x + b X + c Y)^2,
2 b U (-d + x) (c x - a y) + U (b^2 + (d - x)^2) X (c x - a y) -
2 a V x (c (-d + x) + b y) + V (a^2 + x^2) X (c (-d + x) + b y) +
2 U (-d + x) X (a b x + c^2 x - a c y + b c y) Y +
2 V x X (c (c (-d + x) + b y) - a (b (d - x) + c y)) Y +
V (2 a c (c (d - x) - b y) + a^2 (b (d - x) + c y) +
x^2 (b (d - x) + c y)) Y +
U (a (b^2 x + (d - x)^2 x - 2 b c y) +
c (2 b c x + b^2 y + (d - x)^2 y)) Y +
2 U (-d + x) (a c x - b c x + a b y + c^2 y) Y^2 +
2 V x (b c (d - x) + a c (-d + x) + a b y + c^2 y) Y^2 +
V X (a^2 (c (d - x) - b y) + c^2 (c (-d + x) + b y) -
2 a c (b (d - x) + c y)) Y^2 +
U X (c (-b^2 x + c^2 x + 2 b c y) +
a (2 b c x + b^2 y - c^2 y)) Y^2 +
V (2 a c (c (-d + x) + b y) - a^2 (b (d - x) + c y) +
c^2 (b (d - x) + c y)) Y^3 +
U (a (-b^2 x + c^2 x + 2 b c y) +
c (-2 b c x - b^2 y + c^2 y)) Y^3 == 0, X^2 + Y^2 == 0}
*)
