I'm after a closed-form expression for the zeroes of the following function $$ p(z) = d_1 d_2 + d_1\cos(k_1 z) + d_2\cos(k_2 z) $$
$d_1$, $d_2$, $k_1$ and $k_2$ are all real constants. I'm after the values of $z=z_0$ where $p(z_0)=0$.
Does anyone know if there even exists a closed form solution, let alone how to find it?
Thanks.
Suppose $k_2/k_1$ is rational. Then you can write $k_1 = m_1 r$ and $k_2 = m_2 r$ for some $r$, with $m_1$ and $m_2$ integers. If $t = e^{i r z}$, you want to solve $$ d_1 d_2 + \dfrac{d_1}{2} (t^{m_1} + t^{-m_1}) + \dfrac{d_2}{2} (t^{m_2} + t^{-m_2})$$ Multiply by $t^{\max(|m_1|, |m_2|)}$ and you have a polynomial to solve for $t$. It is unlikely that the roots of the polynomial can be expressed in closed form.
EDIT: By using the variable $s = \cos(rz)$ instead, you can halve the degree of the polynomial to solve: $$ d_1 d_2 + d_1 U_{m_1}(s) + d_2 U_{m_2}(s) = 0$$ where $U_{m_i}$ are Chebyshev polynomials. This allows solutions in radicals if $\max(|m_1|,|m_2|) \le 4$.