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I did the same thing you did at first and I could not understand it. But then I realized that the continuous time case was adding $2 \pi$ to $t$, not $w_0$.


The closed contour $C$ must lie inside the region of convergence, so for the ROC $|z|<2$ you have no poles inside $C$ for $n\ge 0$, hence $x[n]=0$ for $n\ge 0$.


HINT: It is based on the fact that $$\sin(x) = \frac{e^{jx} - e^{-jx}}{2j}\quad\text{and}\quad j^2 = -1$$


It's because $$ j \cdot ( a + b) = -\frac{a + b}{j} $$ which stems from the fact that the imaginary unit $j$ has the property : $$ j = \frac{-1}{j} $$

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