I am trying to find out when a linear combination of $\sin(ax)$ and $\cos(bx)$ has an axis of symmetry.
Clearly, $\sin(x)+\cos(x)$ has an axis of symmetry at $\pi/4$. It seems as if $\sin(3 x)+\cos(x)$ does not have an axis of symmetry (could not prove it, but a plot suggests it).
My question is: Is it possible to claim some conditions on the parameters $a$ and $b$ guaranteeing that a linear combination of $\sin(a x)$ and $\cos(b x)$ has an axis of symmetry?
Lets see what happens for $\sin(ax)+cos(x), a\neq1$. for your example, asking for an axis of symmetry is the same as asking for $p \in \mathbb{R}$ such that
$$ \sin(ax+ap) - \sin(-ax + ap) + \cos(x + p ) - \cos(-x + p) = 0 \quad \forall x \in \mathbb{R} $$
we can differentiate this as many times as we want, dropping the constants multiplicatively and obtaining more equations, lets differentiate $4n$ times:
$$ a^{4n}\left( \sin(ax+ap) - \sin(-ax + ap) \right) + \cos(x + p ) - \cos(-x + p) = 0 \quad \forall x \in \mathbb{R} $$
With this process it is easy to prove that the only way to keep this equality true for all $n \in \mathbb{N},x \in \mathbb{R}$ is that each pair of terms be equal to $0$, thus we need $p$ such that
$$ sin(ax + ap) = sin(-ax + ap), \quad cos(x + p) = cos(-x + p) \quad \forall x \in \mathbb{R}$$
From here it follows that for the sine, $p \in \{ \frac{\pi}{2a} + \frac{\pi k}{a}, k\in \mathbb{N} \}$ and the cosine $p \in \{ \pi k, k \in \mathbb{N} \}$. Just to have fun we can search all $a$'s such that there is an axis of symmetry, after some manipulations you can see there is one if and only if $a \neq 1$ can be written as a fraction with an odd numerator and even denominator.
For multiple terms, you can separate each different scale and use the same criteria, most generally I would not hope to find symmetry axes unless choosing too carefully the constants.