Since the numbers and are different by , and since sine is periodic with period , we have also that
Since , we have that lies in the interval
Since you want to solve for , we have by the transition formula above that we might just as well solve
for , where the last equality follows from sine being an odd function.
Since , we have that lies in the interval . Hence both and are numbers in the interval , on which sine is injective, so in solving
for , we can apply to both sides to obtain