= Solution
Let $v_i$ denote the rate of reaction $i$. For <odd oxygen> defined as O plus O$_3$, reactions 5 and 6 each consume one unit, while NO$_2$ <photodissociation> in reaction 7 restores one. Thus the loss from reactions 4–8 is $L=v_5+v_6-v_7$. HO$_2$ stationarity gives $v_5=v_6+v_8+v_4$, and NO$_2$ stationarity within this restricted recycling loop gives $v_7=v_8$. Therefore
$$
L=2v_6+v_4\simeq\boxed{2k_6[\mathrm{HO_2}][\mathrm O]}.
$$
The requested expression neglects the small termination correction $v_4$ in the odd-oxygen budget, while retaining it to determine the much slower total-radical balance. This is consistent when cycling greatly exceeds termination. The net main catalytic cycle is O plus O$_3$ forming two O$_2$ <molecules>; the NO-assisted route mainly interconverts O and O$_3$ and cancels from this restricted net loss after NO$_2$ recycling.
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