# Zeros of $\frac{1}{B(xi)^{1/2}}((iA)^{ix})(ix)^{ix}+ \frac{1}{B(-xi)^{1/2}}((-iA)^{-ix})(-ix)^{-ix}=H(x)$

What would be the zeros of the following function?

$$\frac{1}{B(xi)^{1/2}}((iA)^{ix})(ix)^{ix}+ \frac{1}{B(-xi)^{1/2}}((-iA)^{-ix})(-ix)^{-ix}=H(x)$$

This function is real and I believe it is equal to the cosine of a certain function

$$cos(f(x))=H(x)$$ but what is the function $f(x)$?

The roots satisfy the equation $f(x)=\pi (n+\frac{1}{2})$ and the zeros of $H(x)$ are $x(n)= f^{-1}(n\pi + \frac{\pi}{2})$. $A$ and $B$ are real constants.

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I take it $x$ is real? –  Gerry Myerson Dec 31 '12 at 22:54
yes 'x' is real i search for the real zeros only :) –  Jose Garcia Dec 31 '12 at 23:05