Vrftsjtryk

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

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,
$$
asqrt{x}+bx-pileft(n-frac{1}{2}right)=0
$$
that can be solved with the substitution $x=y^2$ and $xge 0$. 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
$$
by^5+ay-pileft(n-frac{1}{2}right)=0
$$
with no elementary solution.

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