= Solution
Set $\ell=\ln2$. Applying the <exponential series> and the <Maclaurin series> of the <sine> gives
$$
2^x-2^{-x}=e^{\ell x}-e^{-\ell x}=2\ell x+\frac{\ell^3}{3}x^3+O(x^5),\qquad \sin x=x-\frac{x^3}{6}+O(x^5).
$$
Multiply the two <power series>, retaining every contribution through degree four:
$$
(2^x-2^{-x})\sin x=\boxed{2(\ln2)x^2+\frac{(\ln2)^3-\ln2}{3}x^4+O(x^6)}.
$$
Since $0<\ln2<1$, the coefficient of $x^4$ is nonzero, so these are indeed the first two nonzero terms. The remainder is in <Big O notation>.
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