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Swapnil
Swapnil is offline
#5
Dec30-06, 04:01 AM
P: 460
Quote Quote by Chen View Post
Thanks for your tips. I was able to get the equation to the form:
[tex]a^2 cos(ax) + b^2 cos(bx) = c[/tex]
where a,b,c are different constants, but still irrational. Any ideas from here? :)

Thanks!
I spent half-an-hour trying to solve it for x. But I couldn't. I am starting to think that there is no closed form solution to this equation. Anyone else share the same thought?