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Math Help - Trig word problem - solving a trig equation.

  1. #1
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    Trig word problem - solving a trig equation.

    I came across a word problem that I can't seem to puzzle through. I know the answer to it, but I can not seem to find out how to get to it.

    "A block is set in motion hanging from a spring and oscillates about it's resting position x=0, according to the function x=\sqrt{3}sin2t+cos2t. For what values of t is the block at it's resting position x=0."

    This is how I've worked it so far.

    0=\sqrt{3}sin2t+cos2t

    -cos2t=\sqrt{3}sin2t

    -\frac{cos2t}{sin2t}=\sqrt{3}

    -cot2t=\sqrt{3}

    tan2t=\frac{1}{\sqrt{3}}

    2t=tan^-^1(\frac{1}{\sqrt{3}})

    2t = 30\deg

    t = 15\deg

    This answer, unfortunately, is not correct. I know that the correct answer is 75 degrees, but I have no clue how to get to that answer.

    Any help would be appreciated. I've been puzzling over this question for a while now.
    Last edited by mr fantastic; March 14th 2011 at 06:38 PM. Reason: Title.
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  2. #2
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    It's because if \displaystyle -\cot{2t} = \sqrt{3} then \displaystyle \tan{2t} = -\frac{1}{\sqrt{3}}...
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    Quote Originally Posted by Prove It View Post
    It's because if \displaystyle -\cot{2t} = \sqrt{3} then \displaystyle \tan{2t} = -\frac{1}{\sqrt{3}}...
    Your completely correct. Not sure why I forgot about the negative sign, as I had it on the paper I was typing from. However, it still turns out as -15 deg, which doesn't work.
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  4. #4
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    There will be a solution for \displaystyle 2t in the second quadrant and the fourth quadrant.
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  5. #5
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    Oh, now I'm even more confused. The solution that the book gives is either 75 \deg, or \frac{5\pi}{12}, which are in quadrant one.
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    Because when you have the values of \displaystyle 2t, you need to divide them by \displaystyle 2 to get the values of \displaystyle t...
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  7. #7
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    Ah, I understand now. I think I had some faulty notions at play. I was approaching the negative degree as when the spring was below the zero, but I guess the motion is always positive around the unit circle, hence the reason I had to get rid of the negative? Still having a hard time wrapping my head around the underlying concept, but it should click in soon.

    Once again, Prove It, you have helped immensely. Thank you.
    Now when I get home I am going to post a problem that has sin as an exponent, and I haven't got a clue. Not sure why this chapter in the book is so difficult for me.
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