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Thread: Inequality

  1. #1
    Super Member dhiab's Avatar
    May 2009


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  2. #2
    Super Member Deadstar's Avatar
    Oct 2007
    Below is a sin graph between -3\pi and  2\pi.
    The horizontal blue line indicates y=0.5.
    The vertical blue lines show where this cuts the x-axis.
    What you want is every point on the x-axis that is outside the two blue boxes since you can see that if x takes any value inside either of the blue boxes, the corresponding y-value is greater than 0.5.

    Now note that \sin(x) = \frac{1}{2} when x = \frac{\pi}{6} and again when x = \pi - \frac{\pi}{6}. This is the boundary of the box on the positive x-axis and tells us that x cant take any values between x = \frac{\pi}{6} and \pi - \frac{\pi}{6} = \frac{5\pi}{6}

    Since sin(x) is 2\pi periodic, if we subtract 2\pi from the above results we get the values on the negative x-axis that x cant take.

    So putting it all together you get that x can take any value in the following range...
    \bigg{[}-3\pi, -\frac{11\pi}{12}\bigg{]}, \bigg{[}-\frac{7\pi}{6}, \frac{\pi}{6}\bigg{]}, \bigg{[}\frac{5\pi}{6}, 2\pi\bigg{]}
    Attached Thumbnails Attached Thumbnails Inequality-untitled.jpg  
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  3. #3
    MHF Contributor Amer's Avatar
    May 2009
    first take all intervals when the sinx is negative so you will have

    \pi\leq x\leq 2\pi
    in the negative
    -\pi\leq x\leq 0
    -3\pi\leq x\leq -2\pi

    now we need all values of x which sinx less than or equal 1/2

    0\leq x\leq\frac{\pi}{6}
    \frac{5\pi}{6}\leq x\leq \pi

    now in the negative
    \frac{-7\pi}{6}\leq x\leq -\pi

    there is still one interval just I will leave it for you take this graph it will help you ....

    I am always late

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