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Math Help - Solution in the interval

  1. #1
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    Solution in the interval

    Can someone please help me with this problem? I am completely lost on this and I haven't been doing so well in this class at all. If someone could explain how to do this I would really appreciate it. Thank you

    Find all solutions in the interval [0, 2π).

    (sin x)(cos x) = 0
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    MHF Contributor MarkFL's Avatar
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    Re: Solution in the interval

    This equation will be true for:

    a) sin(x) = 0

    b) cos(x) = 0

    For what values of x in the given interval are either true?
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    Re: Solution in the interval

    How do I find out? Besides punching in every number between 0 and 2pi?

    Thanks for answering btw.
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: Solution in the interval

    These are among the special angles which you should have memorized.

    For sin(x) = 0, these represent points on the unit circle whose y-coordinate is zero, that is, are on the x-axis. What angles correspond to these points?

    For cos(x) = 0, these represent points on the unit circle whose x-coordinate is zero, that is, are on the y-axis. What angles correspond to these points?
    Thanks from Darion
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    Re: Solution in the interval

    Quote Originally Posted by Darion View Post
    How do I find out? Besides punching in every number between 0 and 2pi?
    You are suppose to know that already.
    \sin(t)=0 if t=0\text{ or }t=\pi~.

    \cos(t)=0 if t=\frac{\pi}{2}\text{ or }t=\frac{3\pi}{2}~.
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    Re: Solution in the interval

    Oh, Ok. So then sin(x) would be 0 and pi while cos(x) would be pi/2 and 3pi/2
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    Re: Solution in the interval

    Yes, that's right.
    Thanks from Darion
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    Re: Solution in the interval

    Thank you so much guys.

    So if it were cos x = sin x, would the answer be pi/4 and 5pi/4?
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    MHF Contributor MarkFL's Avatar
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    Re: Solution in the interval

    Yes, that is correct as well!
    Thanks from Darion
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    Re: Solution in the interval

    Thank you so much. You guys are awesome.

    I have just one last question if you're willing to help

    Find all solutions to the equation.

    cos2x + 2 cos x + 1 = 0



    I assume this is about the same concept as the other two questions. But I don't know how to set it up.
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    Re: Solution in the interval

    Quote Originally Posted by Darion View Post
    Find all solutions to the equation.
    cos2x + 2 cos x + 1 = 0
    Please learn to post in correct notation.
    Is it \cos(2x)+2\cos(x)+1=0 or \cos^2(x)+2\cos(x)+1=0~?
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    Re: Solution in the interval

    cos^2(x)+2cos(x)+1=0
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    Re: Solution in the interval

    Quote Originally Posted by Darion View Post
    cos^2(x)+2cos(x)+1=0
    Then it becomes (\cos(x)+1)^2=0 or \cos(x)=-1.
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  14. #14
    MHF Contributor MarkFL's Avatar
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    Re: Solution in the interval

    Hint: use a double-angle identity for cosine, so that the equation becomes a quadratic in cos(x).

    edit: Nevermind, I interpreted your equation incorrectly.
    Last edited by MarkFL; November 9th 2012 at 12:37 PM.
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  15. #15
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    Re: Solution in the interval

    I'm sorry, I don't know how to use that.
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