Another Inequality

Apr 2008
748
159
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
 
Oct 2009
4,261
1,836
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