Let $X$ be a topological space, and $T$ its category of open sets with the usual Grothendieck topology. Let $T'$ be any sieve of $T$ (a subcategory of $T$ such that if $U$ is in $T'$ then any subset of $U$ is also in $T'$). For example, $T'$ might be the collection of open subsets subordinate to the open subsets in a cover $\mathcal{U}$. Any sheaf on $T$ induces a functor on $T'$ which can be viewed as a sheaf on $T'$ if $T'$ is given the minimal topology (the only covers are the identity maps). This determines a morphism of topoi $f : T \rightarrow T'$, hence a spectral sequence
$H^p(T', R^q f\_\ast F) \Rightarrow H^{p+q}(T, F)$ .
The Cech cohomology of $F$ with respect to some covering family $\mathcal{U}$ is
$H^p(\mathcal{U}, F) = H^p(T', f\_\ast F)$
where $T' = T'(U)$ is the sieve associated to the cover $\mathcal{U}$. The Cech cohomology is then the filtered colimit
$\check{H}^p(T, F) = \varinjlim\_{(T',f)} H^p(T', f\_\ast F)$
taken over the projections $f : T \rightarrow T'$ associated as above to covering families $\mathcal{U}$.
$\check{H}^p(T, F) \rightarrow H^p(T, F)$
from the spectral sequence, and the question is when these induce an isomorphism. If we could somehow eliminate the $R^p f\_\ast F$, $p > 0$, by passing to a "small enough" cover we would have equality. This condition already holds in many cases; the following condition is more general (but I haven't checked carefully that it actually works!):
For every cover $\mathcal{U}$ of $X$, every $U\_1, \ldots, U\_n \in \mathcal{U}$, and every class in $\alpha \in H^p(U\_1 \mathop{\times}\_X \cdots \mathop{\times}\_X U\_n, F)$, $p > 0$, there exists a refinement $\mathcal{U}'$ of $\mathcal{U}$ such that the restriction of $\alpha$ under the map
$H^p(U\_1 \mathop{\times}\_X \cdots \mathop{\times}\_X U\_n, F) \rightarrow H^p(U'\_1 \mathop{\times}\_X \cdots \mathop{\times}\_X U'\_n, F)$
To make sense of this, one must use some convention for the covers $\mathcal{U}$ and $\mathcal{U}'$ to ensure there is a map as above. For example, one could work only with covers indexed by the points of $X$ (a cover is then a collection of neighborhoods of each point of $X$).
A more refined version of the above condition would say that Cech cohomology equals cohomology in degrees at most $q$ if the above condition holds for $p \leq q$. Since it always holds for $p = 0,1$ this implies that
$\check{H}^1(T, F) = H^1(T, F)$
Edit in response to David's comment:
Here is how Cech cohomology computes cohomology of presheaves. Consider any category $T'$. If $F$ is a presheaf of groups on $T'$ then the sheaf cohomology groups of $F$ are the derived functors of the inverse limit for diagrams of shape $T'$. They are also computed as
where $\mathbf{Z}$ is the constant sheaf associated to the integers. Remarkably, in a presheaf category, $\mathbf{Z}$ has a canonical projective resolution associated to any cover of the final presheaf. A cover of the final presheaf is a collection of objects $U$ of $T'$ such that every object of $T'$ has a map to at least one object of $U$. The $i$-th term of this complex is the direct sum, over all choices of $i$ elements $U_1, ..., U_i$ of $U$, of the groups $\mathbf{Z}_{U_1 \times \cdots \times U_i}$. (You can check this is projective by noting it is the extension by $0$ of $\mathbf{Z}$ from the slice category $T' / U_1 \times \cdots \times U_i$ and extension by $0$ preserves projectives (since it has an exact right adjoint) and $\mathbf{Z}$ is projective on the slice category since all higher cohomology of all sheaves vanishes (since it has a final object). It's also easy to check by a direct calculation.)
Denote this complex by $K$. Since this is a projective resolution of $\mathbf{Z}$, $\mathrm{Hom}(K, F)$ computes the cohomology of $F$. But it is also easy to see that this is just the Cech complex of $F$.
