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sol3 = {h2, h1, d, t, b, c} /.FindInstance[(h2^2 ((1 + d*Sin[t]^2)))/(16 \[Pi]^2) b == 1.14 && (h1*h2 (d (Sin[2 t])))/(2*16 \[Pi]^2) c == 0.1 && 
10^4 < b < 10^6 && 10^4 < c < 10^6 && 0 <= d <= 1 && 0 <= t <= \[Pi]/2 && 0.01 < h2 < 1 && 0.1 < h1 < 1, {h2, h1, d, t,b, c}, Reals, 10,WorkingPrecision -> 10,Method -> "ZengDecision" -> True]

I want to find out all the solutions over the range given above for h1, h2, b and c. But here I just asked for 10 solutions it settles down the lower limit of b and c. But if I changed the range for a and b , from $(10^3,10^6)$ to $(10^2,10^6)$, I even lost these solutions, then it shows error. But still the new range covers the old range. So why it fails to give solutions?

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Your expression produces errors (FindInstance::nsmet, ReplaceAll::reps) with v12 on my Mac

$Version

(* "12.0.0 for Mac OS X x86 (64-bit) (April 7, 2019)" *)

However, using exact numbers in the expression provides three solutions -- near the lower bounds for b and c

sol3 = {h2, h1, d, t, b, c} /. FindInstance[
   (h2^2 ((1 + d*Sin[t]^2)))/(16 π^2) b == 57/50 &&
    (h1*h2 (d (Sin[2 t])))/(2*16 π^2) c == 1/10 &&
    10^4 < b < 10^6 && 10^4 < c < 10^6 && 0 <= d <= 1 &&
    0 <= t <= π/2 && 1/100 < h2 < 1 && 1/10 < h1 < 1,
   {h2, h1, d, t, b, c}, Reals, 10,
   WorkingPrecision -> 10,
   Method -> "ZengDecision" -> True]

enter image description here

Extending the range of b and c produces 18 solutions but does not include the original 3 solutions. Again, the found solutions are near the lower bound for b and c. Presumably, the search starts near the lower bounds and is timing out before getting far from the lower bounds.

sol3rev = {h2, h1, d, t, b, c} /. FindInstance[
   (h2^2 ((1 + d*Sin[t]^2)))/(16 π^2) b == 57/50 &&
    (h1*h2 (d (Sin[2 t])))/(2*16 π^2) c == 1/10 &&
    10^2 < b < 10^6 && 10^2 < c < 10^6 && 0 <= d <= 1 &&
    0 <= t <= π/2 && 1/100 < h2 < 1 && 1/10 < h1 < 1,
   {h2, h1, d, t, b, c}, Reals, 20,
   WorkingPrecision -> 10,
   Method -> "ZengDecision" -> True]

enter image description here

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  • $\begingroup$ That is the problem, it settles down at lower bound. Please any suggestions on how to scan for other solutions $\endgroup$
    – Immy Salam
    Commented Aug 27, 2019 at 14:13
  • $\begingroup$ @ImmySalam - break the intervals into small intervals and use FindInstance or Reduce on each of the smaller intervals. This is likely to take a long time for computation. $\endgroup$
    – Bob Hanlon
    Commented Aug 27, 2019 at 14:24

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