= Solution
Apply the product-to-sum identity directly to the stated modulation:
$$
\boxed{A=a\cos(\omega t-kx)+\frac b2\cos[(\omega+\omega_m)t-kx]+\frac b2\cos[(\omega-\omega_m)t-kx].}
$$
The carrier has angular frequency $\omega$ and the two <modulation sidebands> have angular frequencies $\omega\pm\omega_m$, each with amplitude $b/2$.
\b[Propagation qualification.] In the literal formula, all three terms have the same spatial wavenumber. That identity is correct, but a time-only amplitude applied everywhere is not generally a freely propagating wave at speed $c$. For $k=\omega/c$, applying $\partial_t^2-c^2\partial_x^2$ to the stated field gives $A_0''\cos(\omega t-kx)-2\omega A_0'\sin(\omega t-kx)$, which at $x=t=0$ equals $-b\omega_m^2\ne0$ when the modulation is nontrivial.
A physical modulation launched at a boundary and travelling without dispersion has envelope $a+b\cos[\omega_m(t-x/c)]$. The same expansion then has sideband wavenumbers $k_\pm=(\omega\pm\omega_m)/c$, while its value at $x=0$ agrees with the given time-modulated signal. This distinguishes the requested frequency decomposition from an unsupported fixed-wavenumber propagation claim.
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