$\displaystyle sin2x = \frac{-\sqrt3}{2}$

how to i solve this equation?

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- Nov 23rd 2009, 08:35 PM>_<SHY_GUY>_<multiple angle equations
$\displaystyle sin2x = \frac{-\sqrt3}{2}$

how to i solve this equation? - Nov 23rd 2009, 09:53 PMqmechFirst solve...
First solve:

$\displaystyle

sin y = \frac{-\sqrt3}{2}

$

Then solve

$\displaystyle

y = 2x

$ - Nov 23rd 2009, 09:59 PMbigwave
$\displaystyle sin2x = \frac{-\sqrt{3}}{2}$

$\displaystyle 2x = sin^{-1}(\frac{-\sqrt{3}}{2})$

$\displaystyle x= \frac{1}{2}sin^{-1}(\frac{-\sqrt{3}}{2}) = -\frac{\pi}{3}$

check

$\displaystyle sin(2(\frac{-\pi}{3})) = \frac{-\sqrt{3}}{2}$ - Nov 23rd 2009, 09:59 PM>_<SHY_GUY>_<
- Nov 23rd 2009, 10:15 PMbigwave
this is double angle formula...

$\displaystyle sin2X = 2sinXcosX$

to check x plug it back into the original equation....