Originally Posted by

**craig** Sorry another simple inequality that I can't seem to get the hang of.

$\displaystyle |2(1+\sin{x})|<1$

$\displaystyle \frac{-1}{2}<1+\sin{x}<\frac{1}{2}$

$\displaystyle \frac{-3}{2}<\sin{x}<\frac{-1}{2}$

This is where I can get stuck. I can deal with the $\displaystyle \sin{x}<\frac{-1}{2}$ part, this is just $\displaystyle x<\frac{11\pi}{6} + 2n\pi$.

Not sure what to do with $\displaystyle \frac{-3}{2}<\sin{x}$, every value of $\displaystyle \sin{x}$ is greater than $\displaystyle \frac{-3}{2}$ isn't it?

Thanks again