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

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

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- Nov 29th 2009, 02:23 PMzukywich08tough inverse function problem
1.) -8sin x/2= -4 square root 3

i have no idea how to even start it??? - Nov 29th 2009, 02:49 PMpickslides
divide both sides by -8

this should give you something that looks workable. - Nov 29th 2009, 07:50 PMzukywich08
do you know how to start sin2x= -1/2?

- Nov 30th 2009, 03:02 PMpickslides
$\displaystyle \sin(2x) = \frac{-1}{2}$

what domain does this need to be solved over?

making $\displaystyle 2x = u $

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

this gives

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

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

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