Cohomologies with line bundle vs. coherent coefficients I recently learned in a lecture that the derived category of a smooth variety is generated/spanned by (complexes of) locally free sheaves. (Unfortunately I haven't been able to find a more precise statement.)
Motivated by this, I was wondering if in certain cases, knowledge of cohomologies with line bundle coefficients implies knowledge of the cohomologies with arbitrary coherent coefficients.
For example, suppose that $X$ is a smooth quasi-projective variety (over $\mathbb C$) such that $H^i (X, \mathcal L) = 0$ for $i > 0$ and all line bundles $\mathcal L$; and suppose that all vector bundles are decomposable (i.e. as direct sums of line bundles). In particular, all locally free sheaves have no cohomology in positive degree.
Does this imply that $H^i (X, \mathcal F) = 0$ for $i > 0$ and all coherent sheaves on $X$?
 A: Yes, and for this the abelian version of the theorem on the derived category suffices: Since any coherent sheaf has a finite resolution by locally free sheaves, iterated use of the long exact cohomology sequence together with the assumed acyclicity of locally free sheaves shows that all coherent sheaves are acyclic.
Edit I am sorry that my first answer was too short to be comprehensible. From your post, I had - probably mistakenly - thought that you had actually proved the theorem about the derived category that you quoted, and assumed that it would have been the 'usual' way via proving an 'abelian version' of it first. Anyway, let me give details now.
There is an important (absolute) property of schemes, the resolution property or the existence of enough locall free sheaves: 

Definition: A Noetherian scheme $X$ is said to have enough locally free sheaves if any coherent sheaf ${\mathscr G}$ on $X$ admits an epimorphism ${\mathscr F}\twoheadrightarrow {\mathscr G}$ from a locally free sheaf.

Example 1: An affine scheme has enough locally frees as these correspond to projective modules.
Example 2: Any quasi-projective scheme over a Noetherian ring has this property. Interestingly for your problem, you can even choose ${\mathscr F}$ to be a sum of line bundles: see Corollary 5.18 in Hartshorne.
If a scheme has enough locally free sheaves, you may iterate the choice of locally free covers and see that any coherent sheaf admits a (in general unbounded) resolution by locally free sheaves.
Now, suppose your scheme $X$ is regular of finite Krull dimension $n$ (a smooth variety over a field would do, for example) and that $\ldots\to {\mathscr F}_n\to {\mathscr F}_{n-1}\to\ldots\to{\mathscr F}_0\to {\mathscr G}\to 0$ is a locally free resolution of the coherent sheaf ${\mathscr G}$. Then the kernel ${\mathscr K} := \text{ker}\left({\mathscr F}_n\to {\mathscr F}_{n-1}\right)$ is a locally free sheaf, too: Since this is a local property, it suffices to check that all stalks are free, and this follows since the local rings ${\mathscr O}_{X,x}$ are regular local of dimension $\leq n$, hence of gobal dimension $\leq n$. Hence, truncating the resolution after step $n$ yields a finite resolution of ${\mathscr G}$ by locally free sheaves.
Now, if you have any cohomological functor ${\mathsf T}^{\ast}$ on an abelian category (i.e. a family of functors equipped with connecting homomorphisms turning short exact sequences into long exact sequences - derived functors like sheaf cohomology, for example), then the category of objects ${\mathscr G}$ such that ${\mathsf T}^i(X)=0$ for $i>0$ - the ${\mathsf T}$-acyclic objects - is closed under cokernels. Iterating, it follows that if you have a finite resolution $0\to {\mathscr F}_n\to\ldots\to{\mathscr F}_0\to {\mathscr G}\to 0$ such that all ${\mathscr F}$ are ${\mathsf T}$-acyclic, then ${\mathscr G}$ is ${\mathsf T}$-acyclic, too.
Applying this to your situation, you indeed get the vanishing of all cohomology of line bundles suffices to show the vanishing of all cohomology (and hence your variety is already affine, by Serre's criterion).
