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I got \[y=0, x=2\pi n\]Correct?

seems correct.

But are these the only solutions?

they are already infinite....and u know y must be 0 cos sin of imaginary number can't be real

\[\sin(x+iy)=\sin x \cosh y+i \cos x \sinh y\]

*because sin of...

Cosine can- cosh is a real function. It's imaginary for reals, however

cosine can never be imaginary
hyperbolic cos can be...i think this is your point...

Yes. Thanks, I've got it now.