tough inverse function problem

1.) -8sin x/2= -4 square root 3

i have no idea how to even start it???

divide both sides by -8

this should give you something that looks workable.

do you know how to start sin2x= -1/2?

$\displaystyle \sin(2x) = \frac{-1}{2}$

what domain does this need to be solved over?

making $\displaystyle 2x = u $

$\displaystyle \sin(u) = \frac{-1}{2}$

this gives

$\displaystyle u = \pi+\frac{\pi}{6}, 2\pi-\frac{\pi}{6} = \frac{7\pi}{6}, \frac{11\pi}{6}$

$\displaystyle 2x = \frac{7\pi}{6}, \frac{11\pi}{6}$

$\displaystyle x = \frac{7\pi}{12}, \frac{11\pi}{12}$