Solving an Equation with Sines and Exponents

Date: 05/24/2000 at 04:17:28
From: Serge Boisse
Subject: Solving a trig equation

I am desperately trying to solve the following equation:

     sin(exp(c-x))^2 + sin(exp(c+x))^2 = 0

where x is a (real) unknown and c is a real positive constant. Of 
course, it can be reduced to a system of two simultaneous equations:

     sin(exp(c-x)) = 0
     sin(exp(c+x)) = 0

But that's where I'm stuck; each separate equation has an infinite 
number of solutions. The problem is symmetric, that is, if x is a 
solution, -x is also a solution.

I know that for a lot of values of c, there are no solutions. But for 
some c, there are indeed (a finite number of) solutions. Is there a 
general way of finding these?

Please help.

Date: 05/24/2000 at 08:43:58
From: Doctor Jerry
Subject: Re: Solving a trig equation

Hi Serge,

I can understand why you considered the system:

     sin(e^(c-x)) = 0
     sin(e^(c+x)) = 0

This system can be written as:

     sin(a*e^(-x)) = 0
     sin(a*e^x)    = 0

where a = e^c is a constant, at least as large as 1 (because c is 
positive).

So:
        a*e^x = k1*pi
and
     a*e^(-x) = k2*pi,

where k1 and k2 are integers (positive, negative, or 0).

Letting e^x=y,

     a*y = k1*pi   ...........[1]
     a/y = k2*pi   ...........[2]

So, eliminating y:

     a^2 = k1*k2*pi^2

a >= 1 is known. If integers k1 and k2 can be found so that this 
equation holds, then you can find a solution. Suppose, for example, 
that a = 6*pi. Then we can take k1 = 4 and k2 = 9. From [1]:

     6*pi*y = 4*pi, i.e. y = 2/3

from [2]:

     6*pi = y*9*pi, i.e. y = 2/3

so, from e^x = 2/3, we find x = ln(2/3).

- Doctor Jerry, The Math Forum
  http://mathforum.org/dr.math/