# Another Inequality

#### 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

#### tonio

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

#### craig

you have to take the intersection of the solutions of both inequalities ...

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