Hello I would like to make a Poincare map solving 4 no linear differential equations, I saw a tutorial in mathematica which im doing the same steps, but I want to solve this equations with the restriction that $ \phi 1 (t) = 2 \pi $ . When I use "Mod[]" it only gives me one point, and of course I know many points exists that are solution when $\phi$ is $2 \pi, 4\pi ,6\pi$ and so on.

In[11]:= ec21 = 
  j1'[t] - 2 \[Alpha] j1[t] j2[t] Sin[2 \[Phi]1[t] - 2 \[Phi]2[t]] - 
    2 \[Beta] j1[t] j2[t]^(3/2) Sin[2 \[Phi]1[t] - 3 \[Phi]2[t]] == 0;

ec22 = j2'[t] + 
    2 \[Alpha] j1[t] j2[t] Sin[2 \[Phi]1[t] - 2 \[Phi]2[t]] + 
    3 \[Beta] j1[t] j2[t]^(3/2) Sin[2 \[Phi]1[t] - 3 \[Phi]2[t]] == 0;

ec23 = \[Phi]1'[t] - 1 + 2 j1[t] + 
    3 j2[t] - \[Alpha] j2[t] Cos[
      2 \[Phi]1[t] - 2 \[Phi]2[t]] - \[Beta]  j2[t]^(3/2)
      Cos[2 \[Phi]1[t] - 3 \[Phi]2[t]] == 0;

ec24 = \[Phi]2'[t] - 1 + 3 j1[t] - 
    2 j2[t] - \[Alpha] j1[t] Cos[2 \[Phi]1[t] - 2 \[Phi]2[t]] - 
    3/2 \[Beta] j1[t] j2[t]^(1/2) Cos[2 \[Phi]1[t] - 3 \[Phi]2[t]] == 
   0;



In[55]:= data1 = 
 Reap[NDSolve[{ec21, ec22, ec23, ec24, j1[0] == 0.5, 
      j2[0] == 0.5, \[Phi]1[0] == 0, \[Phi]2[0] == 0, 
      WhenEvent[Mod[\[Phi]1[t], 2 Pi] == 0, 
       Sow[{\[Phi]1[t], \[Phi]2[t], j2[t]}]]}, {}, {t, 0, 100}, 
     MaxSteps -> \[Infinity]]][[-1, 1]][[1]]

Out[55]= {-6.28319, 2.17616, 0.503447}

So how can I get more points? not only 1 if a put for example the independent variable t, it gives me many points

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