$\sin x = 2$

Let $x = A + Bi$ such that $sin (x) = \frac{e^{ix} - e^{-ix}}{2i} = 2 \Rightarrow e^{ix} - e^{-ix} = 4i.$ Further substitution and simplification yields:

$e^{i(A+Bi)} - e^{-i(A+Bi)} = e^{Ai}e^{Bi^2} - e^{-Ai}e^{-Bi^2} = e^{-B}e^{Ai} - e^{B}e^{-Ai} = e^{-B}(cos(A) + i \cdot sin(A)) - e^{B}(cos(A) - i \cdot sin(A)) = 4i$ ;

which after matching the real and imaginary components gives:

$e^{-B} cos(A) - e^{B} cos(A) = 0$ and $e^{-B} sin(A) + e^{B} sin(A) = 4$

which produces $A = \frac{\pi}{2}$ and $e^{-B} + e^{B} = 4 \Rightarrow e^{2B} - 4e^{B} + 1 = 0$ . Let $u = e^{B}$ to obtain the quadratic $u^2 - 4u + 1 = 0 \Rightarrow u = \frac{4 \pm \sqrt{4^2 - 4(1)(1)}}{2} = \frac{4 \pm \sqrt{12}}{2} = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}$ , or $B = ln(2 \pm \sqrt{3}).$

Altogether, $x = A + Bi = \frac{\pi}{2} + ln(2 \pm \sqrt{3}) i$ , which leads to $1 + 2 + 2 + 3 = \boxed{8}.$

$\begin{aligned} \sin x & = 2 \\ \frac i2 \left(e^{-xi} - e^{xi}\right) & = 2 \\ e^{-xi} - e^{xi} & = - 4i & \small \color{#3D99F6} \text{Multiply both sides by }e^{ix} \\ 1 - e^{2xi} & = -4 e^{xi}i \\ e^{2xi} -4 e^{xi}i - 4 & = 1 - 4 \\ \left(e^{xi} - 2i\right)^2 & = -3 & \small \color{#3D99F6} \text{Take square root both sides} \\ e^{xi} - 2i & = \pm \sqrt 3 i \\ e^{xi} & = (2 \pm \sqrt 3)\color{#3D99F6} i & \small \color{#3D99F6} \text{Euler's formula: }e^{\theta i} = \cos \theta + i \sin \theta \\ e^{xi} & = (2 \pm \sqrt 3)\color{#3D99F6} e^{\frac \pi 2i} & \small \color{#3D99F6} \text{Take natural log both sides} \\ ix & = \ln (2\pm \sqrt 3) + \frac \pi 2 i \\ \implies x & = \frac \pi 2 - \ln (2 \pm \sqrt 3) i \end{aligned}$

$\implies a+b+c+d = 1+2+2+3 = \boxed{8}$