Firstly, did you mean
$L(\phi, m_0, m_1, m_2) = P_{1}(\phi)^{m_{0}} \cdot P_2(\phi)^{m_{1}} \cdot P_3(\phi)^{m_{2}}$
'cause P02, P11, P20
are not defined?
If so,we see that
FullSimplify@L[\[Phi], m0, m1, m2]
is
$2^{-\text{m0}-\text{m2}} \cos ^4\left(\frac{\phi }{2}\right)^{\text{m1}} \sin ^2(\phi )^{\text{m0}+\text{m2}}$
With this expression you could simply do any operations.
If we go further:
DL[\[Phi]_, m0_, m1_, m2_] :=FullSimplify@D[FullSimplify@L[\[Phi], m0, m1, m2], \[Phi]]
$2^{-\text{m0}-\text{m2}+1} \csc (\phi ) \cos ^4\left(\frac{\phi }{2}\right)^{\text{m1}} \sin ^2(\phi )^{\text{m0}+\text{m2}} (\cos (\phi ) (\text{m0}+\text{m1}+\text{m2})-\text{m1})$
If we run (as in post Gianluca Gorni):
Reduce[DL[\[Phi], m0, m1, m2] == 0 &&
Element[{m0, m1, m2}, PositiveIntegers] &&
0 <= \[Phi] <= 2 \[Pi], \[Phi]]
we get some (not very) complicated result, from which we could learn that DL[\[Phi], m0, m1, m2] == 0
iff:
$\phi=0,\pi,2 \pi$ or
$\phi=\arccos\left(\frac{\text{m1}}{\text{m0}+\text{m1}+\text{m2}}\right)$ or
$\phi=2 \pi -\arccos\left(\frac{\text{m1}}{\text{m0}+\text{m1}+\text{m2}}\right)$
But if we try to substitute directly
DL[\[Phi], m0, m1, m2] /. \[Phi] -> {0, \[Pi], 2 \[Pi]}
we get
{ComplexInfinity, ComplexInfinity, ComplexInfinity}
We need more accurate testing:
Limit[DL[\[Phi], m0, m1, m2], \[Phi] -> #,
Assumptions ->
Element[{m0, m1, m2}, PositiveIntegers]] & /@ {0, \[Pi], 2 \[Pi]}
{0, 0, 0}
so all five solutions are valid.