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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 $\displaystyle -3\pi $ and $\displaystyle 2\pi$.
    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 $\displaystyle \sin(x) = \frac{1}{2}$ when $\displaystyle x = \frac{\pi}{6}$ and again when $\displaystyle x = \pi - \frac{\pi}{6}$. This is the boundary of the box on the positive x-axis and tells us that x cant take any values between $\displaystyle x = \frac{\pi}{6} $ and $\displaystyle \pi - \frac{\pi}{6} = \frac{5\pi}{6}$

    Since sin(x) is $\displaystyle 2\pi$ periodic, if we subtract $\displaystyle 2\pi$ 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...
    $\displaystyle \bigg{[}-3\pi, -\frac{11\pi}{12}\bigg{]}$, $\displaystyle \bigg{[}-\frac{7\pi}{6}, \frac{\pi}{6}\bigg{]}$, $\displaystyle \bigg{[}\frac{5\pi}{6}, 2\pi\bigg{]}$
    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

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

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

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

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

    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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