Context: While trying to solve for the center of rotation and rotation angle of a uniformly scaled and rotated triangle that will touch the sides of a given $\triangle ABC$ as shown in the figure below, I came across the following trigonometric equation for the angle $\phi$ that the line segments joining the center of rotation with the vertices make with the respective sides:
$\sin^3(\phi) = \sin(A - \phi) \sin(B - \phi) \sin(C - \phi) $
where $A,B,C$ are the vertex angles of $\triangle ABC$.
Short of an analytic solution, I resorted to Newton's method to obtain a numerical solution. A very accurate solution was obtained in just $2$ iterations.
I was wondering if there is an analytic solution to this equation. I am aware that one could use the substitution $u = \tan \dfrac{\phi}{2}$ and end up with a polynomial in $u$ of degree $6$, but other than that, is there any other way ?
There is another way to obtain $x$ solution of the equation $$\sin^3(x) - \sin(A - x) \sin(B - x) \sin(C - x)=0$$
Let $C=\pi-(A+B)$ and $t=\sin(x)$. Replace and expand the sines to obtain $$\big[3-2 \cos (A) \cos (A+2 B)-\cos (2 A) \big]t=4 \sin (A) \sin (B) \sin (A+B)\sqrt{1-t^2}$$ $$\tan(x)=\frac t{\sqrt{1-t^2}}=\frac{4 \sin (A) \sin (B) \sin (A+B) } { 3-2 \cos (A) \cos (A+2 B)-\cos (2 A)}$$