Imaginary Trigonometry

From Euler's identity, we have:

$\begin{aligned} e^{ix} & = \color{#3D99F6}{\cos x} + i \sin x \quad \quad \quad \quad \quad \small \color{#3D99F6}{\text{As } \sin^2 x + \cos^2 x = 1} \\ & = \color{#3D99F6}{\pm \sqrt{1-\sin^2 x}} + i \sin x \quad \quad \small \color{#3D99F6}{\text{Since }\sin x = \frac{3}{2}} \\ & = \pm \sqrt{1-\frac{9}{4}} + i\frac{3}{2} \\ & = \frac{\pm \sqrt{-5}}{2} + i\frac{3}{2} \\ & = i \left(\frac{3 \pm \sqrt{5}}{2} \right) \\ & = e^{i\frac{\pi}{2}} \left( \frac{3 \pm \sqrt{5}}{2}\right) \\ \Rightarrow ix & = i \frac{\pi}{2} + \ln \left(\frac{3 \pm \sqrt{5}}{2}\right) \quad \quad \small \color{#3D99F6}{\times (-i) \text{ throughout}} \\ x & = \frac{\pi}{2} - i \ln \left(\frac{3 \pm \sqrt{5}}{2}\right) \end{aligned}$

$\Rightarrow a + b + c = 3+5+2 = \boxed{10}$

With j = $\sqrt{-1}$ ,

Let x = p - j q,

$\sin (p - j q) = \sin p \cos j q - \cos p \sin j q$ = $\frac32$

$\sin p \cosh q - j \cos p \sinh q$ = $\frac32$

Comparing independent variables for both real and imaginary, we have:

$\sin p \cosh q = \frac32$ and $\cos p \sinh q = 0$

For non-real x of non-zero q, $\sinh q \neq 0$ and therefore $\cos p = 0$ i.e. p = $\frac{\pi}{2}$

With p = $\frac{\pi}{2},$ $\sin \frac{\pi}{2} \cosh q = \frac32$

$\implies \dfrac{e^q + e^{-q}}{2} = \frac{3}{2}$

$\implies$ ( $e^q)^2 - 3 (e^q) + 1 = 0$

$\implies e^q = \dfrac{3 + \sqrt{5}}{2}$ {Only + sign is chosen from $\pm$ referring to question!}

$\implies q = \ln (\dfrac{3 + \sqrt{5}}{2})$

x = p - j q = $\frac{\pi}{2}$ - j $\ln (\dfrac{3 + \sqrt{5}}{2})$ {This may not be a principal value.}

But $\sin [$ $\frac{\pi}{2}$ - j $\ln (\dfrac{3 + \sqrt{5}}{2})]$ = $\frac32.$

Therefore, a + b + c = 3 + 5 + 2 = 10

Answer: $\boxed{10}$