Solving an Equation with Sines and ExponentsDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/