Going complex

$\sin{z}=\frac{3}{4}i$

Multiply by $i$ and add $\cos{z}$ to both sides:

$\cos{z} + i\sin{z}= \cos{z}-\frac{3}{4}\quad \implies \quad \cos{z}=e^{iz}+\frac{3}{4}$

Apply the trigonometrical identity:

$\sin^2{z}+\cos^2{z}=1 \quad \implies \quad -\frac{9}{16}+(e^{iz}+\frac{3}{4})^2=1 \quad \implies \quad (e^{iz})^2+ \frac{3}{2}e^{iz}-1=0$

The solutions for the quadratical equation are $e^{iz}=-2$ and $e^{iz}=\frac{1}{2}$ .

$e^{iz}=-2 \quad \implies \quad iz=\ln{(2 \cdot e^{i\pi+2\pi n})} \quad \implies \quad z= (2n+1)\pi-i\ln{2}, \quad n \in \mathbb{Z}$

$e^{iz}=\frac{1}{2} \quad \implies \quad iz=-\ln{(2 \cdot e^{i2\pi n})} \quad \implies \quad z= 2n\pi+i\ln{2}, \quad n \in \mathbb{Z}$

Both solutions can be rewritten in one expression as $\boxed{z= n\pi+i (-1)^n\ln{2}, n \in \mathbb{Z}}$

$\begin{aligned} \color{#3D99F6} \sin z & = \frac 34 i & \small \color{#3D99F6} \text{By Euler's formula: }e^{\theta i} = \cos \theta + i \sin \theta \\ \frac {e^{zi} - e^{-zi}}{2i} & = \frac 34 i & \small \color{#3D99F6} \text{Multiply both sides by }4i \\ 2e^{zi} - 2e^{-zi} & = - 3 & \small \color{#3D99F6} \text{Multiply both sides by }e^{zi} \text{ and rearrange} \\ 2e^{2zi} +3e^{zi} - 2 & = 0 & \small \color{#3D99F6} \text{Solving the quadratic for }e^{zi} \\ \left(2e^{zi} - 1\right) \left(e^{zi} + 2\right) & = 0 \\ \implies e^{zi} & = \begin{cases} \frac 12 \\ - 2 \end{cases} \end{aligned}$

Let $z=x+yi$ , where $x$ and $y$ are real. Then $e^{zi} = e^{i(x+yi)} = e^{-y+xi} = e^{-y}e^{xi} = e^{-y}(\cos x + i \sin x)$ and

$\begin{cases} e^{-y}(\cos x + i \sin x) = \frac 12 (1+0i) & \implies \begin{cases} x = \cos^{-1} 1 = \sin^{-1} 0 = 2n \pi, \ n \in \mathbb Z \\ e^{-y} = \frac 12 \implies y = \ln 2 \end{cases} \\ e^{-y}(\cos x + i \sin x) = 2(-1+0i) & \implies \begin{cases} x = \cos^{-1} -1 = \sin^{-1} 0 = (2n+1) \pi, \ n \in \mathbb Z \\ e^{-y} = 2 \implies y = - \ln 2 \end{cases} \end{cases}$

Therefore, $z = \begin{cases} 2n\pi + i\ln 2 \\ (2n+1)\pi - i \ln 2 \end{cases} \implies z = \boxed{n\pi + i(-1)^n \ln 2}$ .