= Solution
Nitrogen dioxide is consumed in reactions 1 and 2 and regenerated in reaction $-2$, so
$$
\frac{d[\mathrm{NO_2}]}{dt}=-r_1-r_2+r_{-2}.
$$
Using $r_2-r_{-2}\simeq r_1$ from the nitrate-radical <quasi-steady-state approximation> gives
$$
\boxed{\frac{d[\mathrm{NO_2}]}{dt}\simeq-2k_1[\mathrm{NO_2}][\mathrm{O_3}].}
$$
\b[The factor of two is essential:] an initiating event uses one nitrogen dioxide <molecule>, and the ensuing net radical-trapping event uses a second one. The exact nitrogen balance before making this approximation is $d([\mathrm{NO_2}]+[\mathrm{NO_3}]+2[\mathrm{N_2O_5}])/dt=0$.
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