# Math Help - tough inverse function problem

1. ## tough inverse function problem

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

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

2. divide both sides by -8

this should give you something that looks workable.

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

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

what domain does this need to be solved over?

making $2x = u$

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

this gives

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

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

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