# multiple angle equations

• 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 PM
qmech
First solve...
First solve:

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

Then solve

$\displaystyle y = 2x$
• Nov 23rd 2009, 09:59 PM
bigwave
$\displaystyle sin2x = \frac{-\sqrt{3}}{2}$
$\displaystyle 2x = sin^{-1}(\frac{-\sqrt{3}}{2})$
$\displaystyle x= \frac{1}{2}sin^{-1}(\frac{-\sqrt{3}}{2}) = -\frac{\pi}{3}$

check

$\displaystyle sin(2(\frac{-\pi}{3})) = \frac{-\sqrt{3}}{2}$
• Nov 23rd 2009, 09:59 PM
>_<SHY_GUY>_<
Quote:

Originally Posted by qmech
First solve:

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

Then solve

$\displaystyle y = 2x$

hopefully i did it right, but i got $\displaystyle x = \frac{2\pi}{3}$
is there anything else i have to do?
• Nov 23rd 2009, 10:15 PM
bigwave
this is double angle formula...

$\displaystyle sin2X = 2sinXcosX$

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