tough inverse function problem

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Nov 29th 2009, 02:23 PM
zukywich08

tough 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 PM
pickslides

divide both sides by -8

this should give you something that looks workable.
Nov 29th 2009, 07:50 PM
zukywich08

do you know how to start sin2x= -1/2?
Nov 30th 2009, 03:02 PM
pickslides

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

what domain does this need to be solved over?

making $2x = u$

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

this gives

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

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

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

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