Fairly basic trigonometric equation

[intangible]

Acos(A*x)-Bcos(B*x)=0, where A>B

Is there a general solution for an equation of this form?

Thanks,
intangible

 

[Gib Z]

I'm quite inclined to say there are no solutions for that :(

 

[Feldoh]

Acos(A*x)-Bcos(B*x)=0, where A>B

Is there a general solution for an equation of this form?

Thanks,
intangible

Acos(A*x) = Bcos(B*x) -- So we conclude that A = B

 

[Kaan]

How about A=Pi/2

B=3Pi/2

 

[Gib Z]

I really can not be bothered to expand cos (3x), but if you are, I will verify the rest. I highly doubt it works though.

 

[intangible]

Thanks to everyone participating, I think I got sorted it out by inspecting a similar behaving sine and its infinite product.

\sin x = x \prod_{n = 1}^\infty\left(1 - \frac{x^2}{\pi^2 n^2}\right)

If we assume Asin(pi*A*x)=Bsin(pi*B*x) we may easily inspect the existence of trivial roots if there are any. It turns out there are: for every natural number A and B (A>B) there is a trivial root at x=n. Also, if there is a natural number D so that it divides both A and B, then there will be the trivial roots x=n/D.

A \prod_{n=1}^\infty\left (A x - n)(A x + n)\right = B \prod_{n=1}^\infty\left (B x - n)(B x + n)\right

That leaves A-1 unknown roots per cycle if the former is the case, (A-B)/D per cycle for the latter. Unfortunately, the other roots clearly are irrational, perhaps even transcendent.

Regards,
intangible

(How do I mark this thread solved?)