1. multiple angle equations

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

how to i solve this equation?

2. First solve...

First solve:

$

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

Then solve

$

y = 2x
$

3. $sin2x = \frac{-\sqrt{3}}{2}$
$2x = sin^{-1}(\frac{-\sqrt{3}}{2})$
$x= \frac{1}{2}sin^{-1}(\frac{-\sqrt{3}}{2}) = -\frac{\pi}{3}$

check

$sin(2(\frac{-\pi}{3})) = \frac{-\sqrt{3}}{2}$

4. Originally Posted by qmech
First solve:

$

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

Then solve

$

y = 2x
$
hopefully i did it right, but i got $x = \frac{2\pi}{3}$
is there anything else i have to do?

5. this is double angle formula...

$sin2X = 2sinXcosX$

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