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Thread: Rearranging Equations and Trigonometry

  1. #1
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    Rearranging Equations and Trigonometry

    I am stuck on these last three:

    Question #1

    $\displaystyle
    \frac{tan^2+1}{sec(x)csc(x)}
    $

    My Answer:
    -tan^2(x)+1= sec^2(x).
    -sec/csc= tan (x)

    Question#2

    $\displaystyle 3cot^2(x)=1$

    $\displaystyle
    0\leq x \leq 2 \pi
    $

    Solve the following for x

    My Answer:
    -Divide both sides by 3
    - take the sq. root
    -cot(x)=sqrt3
    -tan of sqrt3 = pi/3
    So is pi/3 my answer?

    Question#3.

    Rearrange the following equation to give an expression for dydx in terms of x and y.

    $\displaystyle
    1-3xy^2\frac{dy}{dx}-6x^2+\frac{dy}{dx}=5xy\frac{dy}{dx}-3x+6x^2\frac{dy}{dx}
    $

    My Answer:

    Lost! Where do I begin?!

    TIA as always!
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  2. #2
    MHF Contributor
    skeeter's Avatar
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    North Texas
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    Quote Originally Posted by mvho View Post
    I am stuck on these last three:

    Question #1

    $\displaystyle
    \frac{tan^2+1}{sec(x)csc(x)}
    $

    My Answer:
    -tan^2(x)+1= sec^2(x).
    -sec/csc= tan (x) ... ok

    Question#2

    $\displaystyle 3cot^2(x)=1$

    $\displaystyle
    0\leq x \leq 2 \pi
    $

    Solve the following for x

    My Answer:
    -Divide both sides by 3
    - take the sq. root
    -cot(x)=sqrt3
    -tan of sqrt3 = pi/3
    So is pi/3 my answer?

    $\displaystyle \textcolor{red}{\cot{x} = \pm \frac{1}{\sqrt{3}}}$

    4 solutions.

    Question#3.

    Rearrange the following equation to give an expression for dydx in terms of x and y.

    $\displaystyle
    1-3xy^2\frac{dy}{dx}-6x^2+\frac{dy}{dx}=5xy\frac{dy}{dx}-3x+6x^2\frac{dy}{dx}
    $

    My Answer:

    Lost! Where do I begin?!

    get every term with dy/dx on the same side , then factor out the dy/dx

    .
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  3. #3
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    Thanks skeeter. A few questions,

    For question #2, you said that there were 4 answers? I only see 2? I know that tangent has a periodicity of pi so my 2 answers are:

    pi/3 and 4pi/3? Where are the other 2 solutions? If I add another period, I would get 7pi/3 which is greater than 2pie

    Ok, I have tacked Question#3, but would like feedback on my algebra. This is me attempting to gather like terms ( I am a bit unsure if I am alllowed to minus the whole term $\displaystyle 5xy\frac{dy}{dx}$ but I went ahead with it.... Should I have divided the left side by $\displaystyle 5xy$ instead?

    I hope what I said makes some sense.


    $\displaystyle
    -3xy^2\frac{dy}{dx}+\frac{dy}{dx}-5xy\frac{dy}{dx}-6x^2\frac{dy}{dx}=-1+6x^2-3x
    $

    Now, factor out the dy/dx..

    $\displaystyle
    \frac{dy}{dx}(-3xy^2+1-5xy-6x^2)=-1+6x^2-3x
    $
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  4. #4
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    you're looking for solutions in the interval $\displaystyle [0 , 2\pi]$

    $\displaystyle \cot{x} = \frac{1}{\sqrt{3}}$ at $\displaystyle x = \frac{\pi}{3}$ and $\displaystyle x = \frac{4\pi}{3}$

    $\displaystyle \cot{x} = -\frac{1}{\sqrt{3}}$ at $\displaystyle x = \frac{2\pi}{3}$ and $\displaystyle x = \frac{5\pi}{3}$

    $\displaystyle
    -3xy^2\frac{dy}{dx}+\frac{dy}{dx}-5xy\frac{dy}{dx}-6x^2\frac{dy}{dx}=-1+6x^2-3x
    $

    Now, factor out the dy/dx..

    $\displaystyle
    \frac{dy}{dx}(-3xy^2+1-5xy-6x^2)=-1+6x^2-3x
    $
    finish by dividing both sides by $\displaystyle (-3xy^2+1-5xy-6x^2)$
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  5. #5
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    OMG YES! ARRGGHH!! I took the square root... hence the 4 solutions!

    That last question wasn't so bad with your help!
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