I need to find the continuous function $f(x)$ that satisfies $f(0)=0$ and:
$$\frac{f(\sin(\pi/6))^2}{\sin^4(\pi/6)}=135$$ $$\frac{f(\sin(\pi/4))^2}{\sin^4(\pi/4)}=63$$ $$\frac{f(\sin(\pi/3))^2}{\sin^4(\pi/3)}=39$$ $$\frac{f(\sin(\pi/2))^2}{\sin^4(\pi/2)}=27$$
The RHS integers above have an interesting interpretation in terms of Knoedel numbers. It is also known that $f(x)$ must satisfy:
$$\frac{f(\sin(\pi/6))^2}{\sin^6(\pi/6)}=540$$ $$\frac{f(\sin(\pi/4))^2}{\sin^6(\pi/4)}=126$$ $$\frac{f(\sin(\pi/3))^2}{\sin^6(\pi/3)}=52$$ $$\frac{f(\sin(\pi/2))^2}{\sin^6(\pi/2)}=27$$
And
$$\frac{f(\sin(\pi/6))^2}{\sin^8(\pi/6)}=2160$$ $$\frac{f(\sin(\pi/4))^2}{\sin^8(\pi/4)}=252$$ $$\frac{f(\sin(\pi/3))^2}{\sin^8(\pi/3)}=69.3333$$ $$\frac{f(\sin(\pi/2))^2}{\sin^8(\pi/2)}=27$$
I guess it's also kind of interesting that the digits in the RHS integers always add up to 9 except for the $f(\pi/3)$ case.
$$f(b(t))=\sqrt{36b(t)^2-9b(t)^4} \:\:\:\:; \:\: b(t)=\sin{t}, \:\:\: 0 < t <\pi/2$$