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Math Help - solve cosec^2 2x - cot 2x = 1

  1. #1
    Member Furyan's Avatar
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    solve cosec^2 2x - cot 2x = 1

    Hello

    The question is to solve

    \csc^2 2x - \cot 2x = 1

    I have tried writing \csc^2 2x in terms of \cot^2 2x

    and \cot 2x in terms of \csc 2x

    without success.

    When I caved in and looked at the mark scheme it says to using \csc^2 2x \equiv 1 + \cot^2 2x, which I didn't get. Could someone please give me a clue and or suggest another way.

    Thank you
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  2. #2
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    Re: solve cosec^2 2x - cot 2x = 1

    Quote Originally Posted by Furyan View Post
    Hello

    The question is to solve

    \csc^2 2x - \cot 2x = 1

    I have tried writing \csc^2 2x in terms of \cot^2 2x

    and \cot 2x in terms of \csc 2x

    without success.

    When I caved in and looked at the mark scheme it says to using \csc^2 2x \equiv 1 + \cot^2 2x, which I didn't get. Could someone please give me a clue and or suggest another way.

    Thank you
    \displaystyle \begin{align*} \csc^2{2x} - \cot{2x} &= 1 \\ \frac{1}{\sin^2{2x}} - \frac{\cos{2x}}{\sin{2x}} &= 1 \\ \frac{1}{\sin^2{2x}} - \frac{\sin{2x}\cos{2x}}{\sin^2{2x}} &= 1 \\ \frac{1 - \sin{2x}\cos{2x}}{\sin^2{2x}} &= 1 \\ 1 - \sin{2x}\cos{2x} &= \sin^2{2x} \\ \sin^2{2x} + \cos^2{2x} - \sin{2x}\cos{2x} &= \sin^2{2x} \\ \cos^2{2x} - \sin{2x}\cos{2x} &= 0 \\ \cos{2x}\left( \cos{2x} - \sin{2x} \right) &= 0 \\ \cos{2x} = 0 \textrm{ or } \cos{2x} - \sin{2x} = 0  \end{align*}

    Solve each of these equations
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  3. #3
    Senior Member MaxJasper's Avatar
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    Re: solve cosec^2 2x - cot 2x = 1

    8 real roots between 0-2pi are:

    \left\{\left\{x\to \frac{\pi }{4}\right\},\left\{x\to \frac{3 \pi }{4}\right\},\left\{x\to \frac{5 \pi }{4}\right\},\left\{x\to \frac{7 \pi }{4}\right\},\left\{x\to \frac{9 \pi }{8}\right\},\left\{x\to \frac{\pi }{8}\right\},\left\{x\to \frac{5 \pi }{8}\right\},\left\{x\to \frac{13 \pi }{8}\right\}\right\}

    Attached Thumbnails Attached Thumbnails solve cosec^2 2x - cot 2x = 1-csc-2x-.png  
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    Re: solve cosec^2 2x - cot 2x = 1

    Substitute 1+cot^2(2x) into the equation

    Then we'll obtain cot^2(2x)-cot(2x)=0

    Factorise the equation
    (cot (2x))(cot (2x)-1)=0

    cot 2x=0
    1/ tan (2x)=0
    For 1/tan (2x)=0,
    tan (2x)= infinity
    So 2x= 90, 270
    x= 45, 135

    cot (2x)-1=0
    1/ tan (2x)=1
    tan (2x)=1
    2x=45, 225
    x=22.5, 112.5
    Thanks from Furyan
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  5. #5
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    Re: solve cosec^2 2x - cot 2x = 1

    Quote Originally Posted by SkyCapri View Post
    Substitute 1+cot^2(2x) into the equation

    Then we'll obtain cot^2(2x)-cot(2x)=0

    Factorise the equation
    (cot (2x))(cot (2x)-1)=0

    cot 2x=0
    1/ tan (2x)=0
    For 1/tan (2x)=0,
    tan (2x)= infinity
    So 2x= 90, 270
    x= 45, 135

    cot (2x)-1=0
    1/ tan (2x)=1
    tan (2x)=1
    2x=45, 225
    x=22.5, 112.5
    Don't forget that the cotangent function is periodic every \displaystyle \begin{align*} 180^{\circ} \end{align*}, so you also need to add all integer multiples of this.
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  6. #6
    Member Furyan's Avatar
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    Re: solve cosec^2 2x - cot 2x = 1

    Quote Originally Posted by Prove It View Post
    \displaystyle \begin{align*} \csc^2{2x} - \cot{2x} &= 1 \\ \frac{1}{\sin^2{2x}} - \frac{\cos{2x}}{\sin{2x}} &= 1 \\ \frac{1}{\sin^2{2x}} - \frac{\sin{2x}\cos{2x}}{\sin^2{2x}} &= 1 \\ \frac{1 - \sin{2x}\cos{2x}}{\sin^2{2x}} &= 1 \\ 1 - \sin{2x}\cos{2x} &= \sin^2{2x} \\ \sin^2{2x} + \cos^2{2x} - \sin{2x}\cos{2x} &= \sin^2{2x} \\ \cos^2{2x} - \sin{2x}\cos{2x} &= 0 \\ \cos{2x}\left( \cos{2x} - \sin{2x} \right) &= 0 \\ \cos{2x} = 0 \textrm{ or } \cos{2x} - \sin{2x} = 0  \end{align*}

    Solve each of these equations
    Thank you very much indeed. I had done exactly that.
    Although I hadn't worked through to the solutions because I had it in my head that I had to get a quadratic in terms of a single trigonometric function.

    It didn't occur to me that I could use R\cos(\theta + \alpha).
    It's good to know there's another approach.

    Although I also now see why \csc^2 2x \equiv \1 + cot^2 2x

    Dividing the identity \sin^2 2x + \cos^2 2x = 1 by  \sin^2 2x.

    I haven't actually worked through the problem yet, but hopefully I'll be able to now. I'll try both ways.

    Thanks again
    Last edited by Furyan; August 24th 2012 at 04:32 AM.
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