1.) -8sin x/2= -4 square root 3
i have no idea how to even start it???
$\displaystyle \sin(2x) = \frac{-1}{2}$
what domain does this need to be solved over?
making $\displaystyle 2x = u $
$\displaystyle \sin(u) = \frac{-1}{2}$
this gives
$\displaystyle u = \pi+\frac{\pi}{6}, 2\pi-\frac{\pi}{6} = \frac{7\pi}{6}, \frac{11\pi}{6}$
$\displaystyle 2x = \frac{7\pi}{6}, \frac{11\pi}{6}$
$\displaystyle x = \frac{7\pi}{12}, \frac{11\pi}{12}$