1. ## Another Inequality

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

2. 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

Yes, exactly, and since you've actually a double inequality (I call it an "and inequality"), you have to take the intersection of the solutions of both inequalities ...

Tonio

3. Originally Posted by tonio
you have to take the intersection of the solutions of both inequalities ...

Tonio
Sorry I'm not sure what you mean by that.