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As per title, I would like to find the zeros of $$ f(x) = \cos(ax^c + bx)$$ where $0\leq x \leq K$, $a \in \mathbb R$, $b \in \mathbb R$, and $c \in (0, 2]$.

I have that $$ f(x) = 0 \Leftrightarrow ax^c + bx = \pi \left(n - \frac{1}{2} \right) $$ Now, since I am not aware of any analytical methods (except for the cases $c=1$, $c=2$, obviously) for solving equations like $$ ax^c + bx - \pi \left(n - \frac{1}{2} \right) = 0 $$ the only method I can think of is to solve this last equation numerically for each choice of $n$ until the solutions reach outside $[0, K]$. Am I correct in assuming there is not a more efficient method? I have considered various things like series expansions, transformations of the variables etc, but without any luck.

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I think that your strategy is a proper one. This can be seen by exploiting some simple cases beyond the trivial ones with $c=1,2$. E.g., let us consider the cases $c=\frac{1}{2},\frac{1}{3},\frac{1}{5}$. So, $$ a\sqrt{x}+bx-\pi\left(n-\frac{1}{2}\right)=0 $$ that can be solved with the substitution $x=y^2$ and $x\ge 0$. You will have an elementary solution. The same can be done for $c=\frac{1}{3}$, provided you will use the Cardano formula. Things worsens for $c=\frac{1}{5}$. Here you can only find a solution by numerical methods. In this case you will have to solve the equation $$ by^5+ay-\pi\left(n-\frac{1}{2}\right)=0 $$ with no elementary solution.

Therefore, aside from very simple particular solutions, you will generally find elementarily unsolvable algebraic equations whose only solution is numerical.

Jon
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