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Math Help - Trigonometric Inequality Problem

  1. #1
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    Trigonometric Inequality Problem

    Hi, this trigonometric inequality was on our test, and I didn't know how to solve it.

    Solve:

    sin(pi/2(x)) < (x-2)^3-1

    The cycle for hte sine function is 1, and I know one of the points of intersection is 3, but I do not know how to solve for the other two points of intersection for the inequality.

    Your help is greatly appreciated!
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  2. #2
    Super Member bigwave's Avatar
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    Quote Originally Posted by KelvinScale View Post
    Hi, this trigonometric inequality was on our test, and I didn't know how to solve it.

    Solve:

    sin(pi/2(x)) < (x-2)^3-1

    The cycle for hte sine function is 1, and I know one of the points of intersection is 3, but I do not know how to solve for the other two points of intersection for the inequality.

    Your help is greatly appreciated!
    is this correct?

    \sin\left(\frac{\pi}{2x}\right) < (x-2)^3-1
    Last edited by bigwave; January 21st 2010 at 03:59 PM.
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  3. #3
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    No it's (pi/2)(x) with the x on top, not the bottom.
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  4. #4
    Super Member bigwave's Avatar
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    Quote Originally Posted by KelvinScale View Post
    No it's (pi/2)(x) with the x on top, not the bottom.
    like this..

    \sin\left(\frac{\pi x}{2}\right) < (x-2)^3-1
    Last edited by bigwave; January 21st 2010 at 04:15 PM. Reason: correct equation
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  5. #5
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    yes, I think you can solve it by guess and check, but I was wondering if there was a systematic way of doing it.
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  6. #6
    Super Member bigwave's Avatar
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    my assumption on this. is that since it is an inequality you have 2 graphs
    and the answer is a particular area between the two

    you might want to look at this.....
    make sure I have plugged it in correct

    http://www.wolframalpha.com/input/?i=+sin((x*pi)%2F2)+%3C+(x-2)^3-1

    hope this helps...
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by bigwave View Post
    my assumption on this. is that since it is an inequality you have 2 graphs
    and the answer is a particular area between the two

    you might want to look at this.....
    make sure I have plugged it in correct

    http://www.wolframalpha.com/input/?i=+sin((x*pi)%2F2)+%3C+(x-2)^3-1

    hope this helps...
    the answer has nothing to do with the area between the two. From the graph we can see that for (approximately) x > 2.587, the cubic exceeds the sine function, and hence is greater on that interval. It looks like it would be pretty annoying to find the exact value for the left end-point of the interval (if there even is one!), to find this value, you can use some kind of approximation method, like the Newton-Raphson method--if you're a glutton for punishment.
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