What is the solution of $\sin z=\cosh 4$?
By putting $z=x+iy$ I managed to find that the real part of $z$ is $x= \frac \pi 2+2n\pi $, but the imaginary part is contradictory giving negative value of $\cosh x$.
The answer is given to be $\frac \pi 2+2n\pi \pm \cosh 4$.
On
Well, $\cosh 4$ is just a number, so we apply inverse functions... \begin{align*} \sin z &= \cosh 4 \\ z = \sin^{-1} \cosh 4 + 2\pi k &\text{ or } z = \pi - \sin^{-1} \cosh 4 + 2 \pi k, \text{ for any integer $k$}. \end{align*} Suppose we already know $\sin^{-1}\cosh 4 = \frac{\pi}{2} - 4\mathrm{i}$. Then $$ z = \frac{\pi}{2} - 4\mathrm{i} + 2\pi k \text{ or } z = \frac{\pi}{2} + 4\mathrm{i} + 2\pi k, \text{ for any integer $k$}. $$ We could perhaps simplify this to $z = \frac{\pi}{2} \pm 4\mathrm{i} + 2\pi k, \text{ for any integer $k$}$. This is as close as I can get any answer to the form you gave (since $\cosh 4 = 27.308\dots \neq 4\mathrm{i}$).
The equation can be written $$ e^{iz}-e^{-iz}=2i\cosh 4 $$ or $$ e^{2iz}-2ie^{iz}\cosh 4-1=0 $$ Set $w=e^{iz}$, and solve the quadratic $w^2-2iw\cosh 4-1=0$: $$ w=i\cosh4\pm\sqrt{-\cosh^24+1}=i(\cosh4\pm \sinh4) $$ Thus the roots are $$ ie^4,\quad ie^{-4} $$ Now, if $z=x+iy$, the first root gives $$ e^{-y}e^{ix}=ie^{4}= e^4\left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right) $$ so $$ z=\frac{\pi}{2}+2k\pi-4i $$ Similarly, the second root gives the solutions $$ z=\frac{\pi}{2}+2k\pi+4i $$
If you want to go the $\sin(x+iy)$ way, consider $$ \sin x\cos(iy)+\cos x\sin(iy)=\cosh4 $$ that can be rewritten using $\cos(iy)=\cosh y$ and $\sin(iy)=i\sinh y$: $$ \sin x\cosh y+i\cos x\sinh y=\cosh 4 $$ Thus you have $$ \sin x\cosh y=\cosh 4, \qquad \cos x\sinh y=0 $$ From the second equation you get $\cos x=0$ or $y=0$; the case $y=0$ is incompatible with the first equation, because $\cosh4>1$. So we have $$ x=\frac{\pi}{2}+2k\pi \qquad\text{or}\qquad x=-\frac{\pi}{2}+2k\pi $$ but the second set of solution must be excluded, because it would give the incompatible relation $-\cosh y=\cosh4$. So you remain with the first set of solutions, from which $$ \cosh y=\cosh4 $$ and so $y=\pm4$.