# Solving $\frac{\sin(2v)+\sin(40^\circ)}{\sin(40^\circ-2v)}=3$

How do I solve this equation?

$$\frac{\sin(2v)+\sin(40^\circ)}{\sin(40^\circ-2v)}=3$$

I'm not sure that I have the knowledge required to solve this problem yet, I have tried and can get it to: $$\sin(2v)(1+3\cos(40^\circ))=3\sin(40^\circ)\cos(2v)-\sin(40^\circ)$$

However, I doubt it has brought me any closer to a solution, and would be happy if you could provide me with a full solution. Thanks!

HINT:

Using Prosthaphaeresis Formula $(\sin C+\sin D)$,

$$2\sin(v+20^\circ)\cos(v-20^\circ)=-3\cdot2\sin(v-20^\circ)\cos(v-20^\circ)$$

$$2\cos(v-20^\circ)\{\sin(v+20^\circ)+3\sin(v-20^\circ)\}=0$$

So, either $\cos(v-20^\circ)=0$ or $\sin(v+20^\circ)+3\sin(v-20^\circ)=0$

But $\cos(v-20^\circ)=0$ will make the left hand side of the given expression $$\dfrac00$$

Finally use $$\sin(A\pm B)$$ formula