Finding solutions for $\sin(z)=\sin(2)$ having some difficulty wrapping my head around how to methodically do this style of question:
Find all solutions $z\in ℂ$ for $\sin(z)=\sin(2)$
I attempted to solve this by using the identity,
$\sin(x)\cosh(y)+i\cos(x)\sinh(y)=\sin(2)$, 
so that,
$\sin(x)\cosh(y)=\sin(2)$ and $\cos(x)\sinh(y)=0$,
but this seemed to lead to a dead end, so I'm unsure as to what the next step should be...
Any help is much appreciated

$$\begin{align}
  & \text{This problem still can be solved following your procedure}: \\ 
 & \cos \left( x \right)\sinh \left( y \right)=0\Rightarrow \cos \left( x \right)=0\ OR\ \sinh \left( y \right)=0 \\ 
 &  \\ 
 & If\sinh \left( y \right)=0,\ then\ y=0\ ,and\ hence\  \\ 
 & \sin \left( x \right)\cosh \left( y \right)=\sin \left( x \right)\cosh \left( 0 \right) \\ 
 & \quad \quad \quad \quad \quad \quad =\sin \left( x \right)\times 1 \\ 
 & \quad \quad \quad \quad \quad \quad =\sin \left( x \right) \\ 
 & \quad \quad \quad \quad \quad \quad \Rightarrow \sin \left( x \right)=\sin \left( 2 \right)\Rightarrow x=2+2n\pi . \\ 
 &  \\ 
 & If\ \cos \left( x \right)=0,\ then\ \sin \left( x \right)=\pm \sqrt{1-\cos {{\left( x \right)}^{2}}}=\pm \sqrt{1-{{0}^{2}}}=\pm 1,and\ hence \\ 
 & \ \sin \left( x \right)\cosh \left( y \right)=\pm \cosh \left( y \right)=\sin \left( 2 \right) \\ 
 & \quad \quad \quad \quad \quad \quad \,\,\Rightarrow \ \cosh \left( y \right)=\pm \sin \left( 2 \right)\  \\ 
 & \quad \quad \quad \quad \quad \quad \,\,\Rightarrow \ \cosh \left( y \right)=\sin \left( 2 \right)\ since\ \cosh \left( y \right),\sin \left( 2 \right)>0.\  \\ 
 & \quad \quad \quad \quad \quad \quad \,\,\Rightarrow y={{\cosh }^{-1}}\left( \sin \left( 2 \right) \right). \\ 
 &  \\ 
 & Finally\ z=\left( 2+2n\pi  \right)+i{{\cosh }^{-1}}\left( \sin \left( 2 \right) \right) \\ 
\end{align}
$$
