Scaled sines equation: $ C \cos(Ax) = \cos(x) $ $$
C \cos(Ax) = \cos(x)
$$
$C, A, x \in \mathbb{R}$. Is there a known solution (for $x$)? If not, how could one approach it? WA struggles.
I thought of solving it for $C$ for each $x$ and for some $A_0$, then for another $A_1$ and seeing if it can generalize. Arrcosining until a full period is tiled seems to be a piecewise nightmare.
My goal is its inequality, which can be derived from the equality by solving periodicity, so if the inequality is somehow easier then that's also welcome.
 A: For the case where $A\in \Bbb Z$, one can use the Tchebyshev polynomial $T_n(x)$ to write the equation as $$C\; T_A(\cos x) = \cos x$$ which is a polynomial equation of degree $A$ in $\cos x$.
It's now clear that in general won't be a closed form for $A\ge 5$, but if some roots can be obtained, then the solutions to the original equation can be obtained from $\cos x = \alpha_i$ where $\alpha_i$ are the roots of the polynomial (Since you are interested only in real solutions, you obviously discard the roots with absolute value greater than $1$).
A: First try $x=\tan^{-1}(y)$
$$c \cos(a \tan^{-1}(y))=\frac1{\sqrt{y^2+1}}$$
Then expand into a power function equation:
$$1/2 c ((1 - i y)^a + (1 + i y)^a) (y^2+1)^{-\frac a2}=\frac 1{\sqrt{y^2+1}} \iff((1 - i y)^a + (1 + i y)^a) (y^2+1)^\frac {1-a}2=\frac2c$$
Now we need the inverse of $((1 - i y)^a + (1 + i y)^a) (y^2+1)^\frac {1-a}2$. If $a\in\Bbb N$, then you expand the equation. There are a few ways to invert a general polynomial, but they require the Riemann theta or other functions. Alternatively, let’s use root notation:
$$x=\tan^{-1}\left(\text{Root}\left[\frac c2((1 - i y)^a + (1 + i y)^a) (y^2+1)^\frac {1-a}2-1,1\right]\right)$$
