# Math Help - help, please and thank you

1. ## help, please and thank you

find exact solutions of 2sin2x = -1, for 0 (<or=)x<2pi

Thanks a bunch

2. Originally Posted by Luckegrl_16
find exact solutions of 2sin2x = -1, for 0 (<or=)x<2pi

Thanks a bunch
$2sin(2x) = -1$

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

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

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

-Dan

3. ## yaya

oh and srry about the whole multiple posts thing, I've never really used these b4, but now I know