= Solution
Use the <complex exponential> forms of the <sine> and <cosine>, and put $w=e^{2iz}$. Multiplying the equation by $2e^{iz}$ gives $(2-i)(w-1)/i+(w+1)=0$, which simplifies to $w=1-i$. Taking all branches of the <complex logarithm>,
$$
2iz=\frac12\ln2+i\left(-\frac\pi4+2\pi n\right).
$$
Thus every solution is
$$
\boxed{z=n\pi-\frac\pi8-\frac{i}{4}\ln2,\qquad n\in\mathbb Z.}
$$
The exponential factor used to multiply the equation is nonzero, so no solutions were introduced or lost.
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