Dear Bill,
Thanks a lot for your help, I see that Reduce does give the answers but I need to know how he calculates it. So I will need the expression of p without the given values to the constants. I have inserted these values only to test and see if my expressions were right. Most of the values do have restrictions though.
0< p>=1(so 2Pi is not even necessary); dq=4Pi(max); da=1.5Pi(max); do=Pi(max)
For example if I have the next formula:
c1sin(c2x)==0
I will need the following answer: x = Pi * (N / c2) So I can program this into the machine. The maximum amount of zero's (so minima and maxima) between 0 and 1 is 6.
What I still don't understand is why I do not get the same answer as you do while I am using the same values and formula:
In[45]:= N[Reduce[Xlinedp[p] == 0 && 0 < p <= 1, p]]
Out[45]= (C[1] [Element] Integers && 0.447508 < C[1] <= 0.697508 && p == 0.31831 (12.5664 C[1] - 5.62355)) ||
(C[1] [Element] Integers && 0.403085 < C[1] <= 0.653085 && p == 0.31831 (12.5664 C[1] - 5.06532)) ||
(C[1] [Element] Integers && 0.346915 < C[1] <= 0.596915 && p == 0.31831 (12.5664 C[1] - 4.35946)) ||
(C[1] [Element] Integers && 0.302492 < C[1] <= 0.552492 && p == 0.31831 (12.5664 C[1] - 3.80123)) ||
(C[1] [Element] Integers && 0.243817 < C[1] <= 0.493817 && p == 0.31831 (12.5664 C[1] - 3.06389)) ||
(C[1] [Element] Integers && 0.221104 < C[1] <= 0.471104 && p == 0.31831 (12.5664 C[1] - 2.77848)) ||
(C[1] [Element] Integers && 0.17487 < C[1] <= 0.42487 && p == 0.31831 (12.5664 C[1] - 2.19749)) ||
(C[1] [Element] Integers && 0.138254 < C[1] <= 0.388254 && p == 0.31831 (12.5664 C[1] - 1.73735)) ||
(C[1] [Element] Integers && 0.0809532 < C[1] <= 0.330953 && p == 0.31831 (12.5664 C[1] - 1.01729)) ||
(C[1] [Element] Integers && 0.0359555 < C[1] <= 0.285956 && p == 0.31831 (12.5664 C[1] - 0.45183)) ||
(C[1] [Element] Integers && -0.0205039 < C[1] <= 0.229496 &&
p == 0.31831 (12.5664 C[1] + 0.257659)) ||
(C[1] [Element] Integers && -0.0629505 < C[1] <= 0.18705 && p == 0.31831 (12.5664 C[1] + 0.791059)) ||
(C[1] [Element] Integers && -0.121684 < C[1] <= 0.128316 && p == 0.31831 (12.5664 C[1] + 1.52912)) ||
(C[1] [Element] Integers && -0.128316 < C[1] <= 0.121684 && p == 0.31831 (12.5664 C[1] + 1.61247)) ||
(C[1] [Element] Integers && -0.18705 < C[1] <= 0.0629505 && p == 0.31831 (12.5664 C[1] + 2.35053)) ||
(C[1] [Element] Integers && -0.229496 < C[1] <= 0.0205039 && p == 0.31831 (12.5664 C[1] + 2.88393)) ||
(C[1] [Element] Integers && -0.285956 < C[1] <= -0.0359555 &&
p == 0.31831 (12.5664 C[1] + 3.59342)) ||
(C[1] [Element] Integers && -0.330953 < C[1] <= -0.0809532 &&
p == 0.31831 (12.5664 C[1] + 4.15888)) ||
(C[1] [Element] Integers && -0.388254 < C[1] <= -0.138254 && p == 0.31831 (12.5664 C[1] + 4.87894)) ||
(C[1] [Element] Integers && -0.42487 < C[1] <= -0.17487 && p == 0.31831 (12.5664 C[1] + 5.33908)) ||
(C[1] [Element] Integers && -0.471104 < C[1] <= -0.221104 && p == 0.31831 (12.5664 C[1] + 5.92007)) ||
(C[1] [Element] Integers && -0.493817 < C[1] <= -0.243817 && p == 0.31831 (12.5664 C[1] + 6.20548))
If it is not possible to find my solutions(constants undefined) with Reduce or Solve, is there any other way I could solve this problem?