Simon,

I doubt very much that you will find what you are looking for in fact I very much expect that there is no solution in terms of elementary functions.

I did experiment with some graphing software and, at least for small values of n, 1≤ n ≤ 20, I saw that the behaviour depended on n. For n = 2, 3, 5, 7, 11, 13, 17 and 19 it looked like there was no solution in (0, 1). For n = 4 and 9 the only solution is (0, 1) seems to be x = 1/2. For n = 6, 8, 10, 14, 15 and 16 it looks like there is one solution in (0, 1/2) and for n = 12, 18 and 20 there seem to be 2 solutions in (0, 1/2).

You didn't say how you want to use this but one possibility is to approximate the function f(x) = cos²(π × n^x) + cos²(π × n^1-x) - 2 using a power series. I am no expert on this and I don't even have a suggestion of the type of series to try.

I'm sorry I'm not much help,
Harley

Well, RD showed me wrong in a very nice way. Harley

Simon,

It's not as hard as you may have thought. Assuming we're looking only at real numbers here, cos²(alpha) cannot be outside [0,1] for any alpha.

(1) So what can you say about

cos²(π × n^x) and cos²(π × n^1-x)?

(2) Then what can you say about

cos(π × n^x) and cos(π × n^1-x) ?

(3) And what does this tell you about n^x and n^1-x?

(4) How are n^x and n^1-x related? (Hint: what is their product?)

(5) You should now be able to see why there are (eg) two solutions in (0, 0.5) for n=12, and to write them as ratios of logarithms.

By the way, are solutions in (0, 0.5) really sufficient? If n is a perfect square, do you really not want the solution x=0.5)?

Good Hunting!
-RD