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If $f$ has a double zero at a point $z$ then not only $z=\sin z$ there but also the derivatives must be the same, so that $1=\cos z$ there. Letting $w=e^{iz}$ we obtain the equation $w+\frac1w=2i$ from which it follows $w=i(1\pm\sqrt2)$ is pure imaginary. But then $z=\frac{w-\frac1w}{2i}$ is real, a contradiction.