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I want to solve transcendental equations by Mathematica 7. Solve does not help me, reduce - as well. Seams that I shoulkd use FindRoot, but it does not give all solutions. It gives only one solution. So how can I solve transcendental equations?

Here is the equation I want to solve:

enter image description here

So for a fixed value of L, A1, A2 (L >= 0 and is an integer) parameters I want to find all values for k expressed with A1 and A2, for which the determinant is 0. In addition I must say that A1 and A2 >0.

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Can you give an example of an equation you want to solve? – Yuval Filmus Feb 20 '11 at 9:02

Finding all solutions of a general transcendental equation is a nontrivial task. As you have more than one equation this even looks hopeless ;-) So unless you know more about the equations and the allowed range for the variables (real, complex, from a compact set, ...), Mathematica (and this forum) cannot help.

For your specific problem, Mathematica can help. Bessel functions of half integer arguments can be reduced to trigonometric functions. Mathematica will help you with this task if you write 1/2 instead of 0.5. So for L=0, you obtain that the determinant is proportional to $\sin[ (A_1 -A_2)k]$ and therefore the solutions are at $k= n \pi/(A_1-A_2)$ with $n\in \mathbb{Z}$. For $L=1,2,…$ the busyness gets more and more tricky, but numerical solutions are always obtainable.

E.g., for $L=1$ the determinant is zero whenever $(1+A_1 A_2 k^2) \sin[(A_1-A_2)k] - (A_1 -A_2) k \cos[(A_1-A_2)k]=0$. There is a trivial solution for $k=0$. For $|k| \geq (A_1 -A_2)/A_1 A_2$ this equation has no solutions. So you only need to find the solutions in a finite interval (except for $A_1 = A_2$ where the determinant is always zero). For this FindRoot helps.

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Not directly on-topic, but Mathematica 8's Solve gives the following solutions

$$\begin{array}{l} \cos \left(A_2 k\right)\to \sin \left(A_2 k\right) \cot \left(A_1 k\right) \\ \cos \left(A_1 k\right)\to 0 \\ \sin \left(A_1 k\right)\to 0 \\ \sin \left(A_2 k\right)\to 0 \\ \sin \left(A_1 k\right)\to 0 \end{array}$$

L = 0;
mat = {{BesselJ[L + 1/2, k A1],BesselJ[-(L + 1/2), k A1]},
      {BesselJ[L + 1/2, k A2], BesselJ[-(L + 1/2), k A2]}};
Solve[Det[mat] == 0]
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