Given $2\sin A + \sin B = 2\sin C$ in a triangle find minimum of $\frac{5}{\sin A} + \frac{9}{\sin C}$ Given $2\sin A + \sin B = 2\sin C $ where $A,B,C$ are angles in a triangle find minimum of $\frac{5}{\sin A} + \frac{9}{\sin C}$
so $\sin C - \sin A = \frac{ \sin B}{2}$ so $2\cos \frac{C+A}{2} \sin \frac{C-A}{2} = \frac{ \sin B}{2} = 2\sin \frac{B}{2} \sin \frac{C-A}{2} $  so $\sin \frac{C-A}{2} = \frac{1}{4}$
I feel like I am awfully close here, but how do I finish it?

In terms of the sides and circumradius:
$$2\sin A + \sin B = 2\sin C\iff \dfrac{a}{R} + \dfrac{b}{2R} = \dfrac{c}{R}\iff 2c=b+2a.$$
Then, you can turn your expression into a function of two variable, say $a,b$, using the same Law of Sines and take derivatives, using the formula:
$$R = \dfrac{abc}{4S},$$
where $S$ is the area.
