= Solution
Use the exponential definitions of <complex sine function> and <complex cosine>. Multiplying the equation by $2i$ gives $e^{iz}-e^{-iz}=-2(e^{iz}+e^{-iz})$, hence $3e^{iz}+e^{-iz}=0$. Since exponentials never vanish, this is equivalent to $e^{2iz}=-1/3$. All branches of the <complex logarithm> give
$$
2iz=-\ln3+i(\pi+2\pi n),\qquad\boxed{z=\frac\pi2+n\pi+\frac i2\ln3,\quad n\in\mathbb Z.}
$$
The derivation does not divide by $\cos z$, so it introduces no lost zero-denominator cases.
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