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Math Help - Trigonometric Equation

  1. #1
    DCB
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    Trigonometric Equation

    Hi everyone,

    I've been given the following equation:


    and been asked to solve it the interval

    I've spent hours basically going in circles, unable to solve it. If anyone could show me how to simplfy this I would really appreciate it.

    Thanks
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  2. #2
    MHF Contributor

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    Re: Trigonometric Equation

    My strategy would be to reduce to a single trig function. First, cot(a)=\frac{cos(a)}{sin(a)} so that cos(2x)= \frac{cos(2x)}{sin(2x)}= \frac{cos^2(x)- sin^2(x)}{2sin(x)cos(x)}= \frac{1- 2sin^2(x)}{2sin(x)\sqrt{1- sin^2(x)}}. Similarly, tan(x)= \frac{sin(x)}{cos(x)}= \frac{sin(x)}{\sqrt{1- sin^2(x)}}. So this equation says
    \frac{3(1- 2sin^2(x))}{2sin(x)\sqrt{1- sin^2(x)}}+ \frac{sin(x)}{\sqrt{1- sin^2(x)}}- 4+ \frac{3}{sin^2(x)}= 0

    "Clear the fractions" by multiplying through by 2sin^2(x)\sqrt{1- sin^2(x)}:
    3sin(x)(1- 2sin^2(x)+ 2sin^3(x)- 8sin^2(x)\sqrt{1- sin^2(x)}+ 6\sqrt{1- sin^2(x)}= 0

    Let y= sin(x) so this becomes 3y(1- 2y^2)+2y^3- 8y^2\sqrt{1- y^2}+ 6\sqrt{1- y^2}= 0

    We can write this as 3y(1- y^2)+ 2y^3= (8y^2+ 6)\sqrt{1- y^2}
    and squaring both sides gives a sixth degree polynomial in y.
    Last edited by HallsofIvy; January 1st 2014 at 05:22 AM.
    Thanks from DCB
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  3. #3
    DCB
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    Re: Trigonometric Equation

    Thanks a lot for your help! It's much appreciated.

    I think you may have made two small errors (either that or I misunderstood something)

    I think (8y2+6)√(1-y2)
    should rightly be (8y2-6)√(1-y2)

    and you also dropped the 2 from 3y(1-2y3) at the end. Unless I made some mistake.

    In any case doing things the way you showed me I ended up with

    16sin6x-88sin4x+105sin2x-36

    which I then factored and got 3 facts, 2 of which were the same, and one of which was impossible (sin(x)=2 when simplified)
    In any case the final answer, if I didn't mess up somewhere along the way, is x=pi/3, 2pi/3, or 4pi/3 in the given interval.

    Thanks a lot for your help, I don't know if I ever would have found the answer without you.
    Oh and sorry, for some reason I am heaving trouble with latex all of a sudden. I'll look into that later though.
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  4. #4
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    Re: Trigonometric Equation

    Could you check that you've copied the equation correctly ?
    If that second term were 2\tan x I think it drops out quite nicely.
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  5. #5
    DCB
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    Re: Trigonometric Equation

    I did indeed copy it correctly. The question from a practice exam for my universities entrance exam and I think it not dropping out nicely is expressly their purpose unfortunately.
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  6. #6
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    Re: Trigonometric Equation

    Quote Originally Posted by DCB View Post
    Thanks a lot for your help! It's much appreciated.

    I think you may have made two small errors (either that or I misunderstood something)

    I think (8y2+6)√(1-y2)
    should rightly be (8y2-6)√(1-y2)

    and you also dropped the 2 from 3y(1-2y3) at the end. Unless I made some mistake.

    In any case doing things the way you showed me I ended up with

    16sin6x-88sin4x+105sin2x-36

    which I then factored and got 3 facts, 2 of which were the same, and one of which was impossible (sin(x)=2 when simplified)
    In any case the final answer, if I didn't mess up somewhere along the way, is x=pi/3, 2pi/3, or 4pi/3 in the given interval.

    Thanks a lot for your help, I don't know if I ever would have found the answer without you.
    Oh and sorry, for some reason I am heaving trouble with latex all of a sudden. I'll look into that later though.
    Just for information Mathematica is showing an additional solution in this interval at x=2\arctan\left(\frac{1}{2}\left(1+\sqrt{5}\right) \right)
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  7. #7
    DCB
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    Re: Trigonometric Equation

    Really? Any idea how I could go abut finding that answer? And also unless I am mistaken isn't that the golden ratio?
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  8. #8
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    Re: Trigonometric Equation

    Trigonometric Equation-02-jan-14.png
    Thanks from MINOANMAN and DCB
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  9. #9
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    Re: Trigonometric Equation

    It certainly is. Nice catch.

    Interesting. Ibdutt's cubic is also showing 4 solutions including x=\pi-\arctan(2) which turns out to be equal to the golden ratio solution but in a nicer form.
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