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Math Help - Another Inequality

  1. #1
    Super Member craig's Avatar
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    Another Inequality

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

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

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

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

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

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

    Thanks again
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    Quote Originally Posted by craig View Post
    Sorry another simple inequality that I can't seem to get the hang of.

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

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

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

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

    Not sure what to do with \frac{-3}{2}<\sin{x}, every value of \sin{x} is greater than \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
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    Super Member craig's Avatar
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    Quote Originally Posted by tonio View Post
    you have to take the intersection of the solutions of both inequalities ...

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