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.