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Math Help - trig equation - all solutions

  1. #1
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    trig equation - all solutions

    solve  cosec^{2}2x - cot2x = 1

      0  \leq x \leq 180


    replace cosex^{2}2x with  1 + cot^{2}2x = cosex^{2}2x

     1 + cot^{2}2x -cot2x = 1

     cot^{2}2x -cot2x = 0

     cot2x(cot2x-1) = 0

     cot2x = 0 or

     cot2x = 1

    for  cot2x =0 is it correct to say there are no solutions for this?

    for the other one I got  tan2x = 1

     2x = 45, 225

     x = 22.5 , 112.5

    Are these all the solutions in the interval, or does cot2x =0 also have a solution?
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  2. #2
    Super Member bigwave's Avatar
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    Quote Originally Posted by Tweety View Post
    solve  cosec^{2}2x - cot2x = 1

      0  \leq x \leq 180


    replace cosex^{2}2x with  1 + cot^{2}2x = cosex^{2}2x

     1 + cot^{2}2x -cot2x = 1

     cot^{2}2x -cot2x = 0

     cot2x(cot2x-1) = 0

     cot2x = 0 or

     cot2x = 1

    for  cot2x =0 is it correct to say there are no solutions for this?

    for the other one I got  tan2x = 1

     2x = 45, 225

     x = 22.5 , 112.5

    Are these all the solutions in the interval, or does cot2x =0 also have a solution?
    on \cot{2x} = 0

    \cot^{-1}{(0)} = \frac{\pi}{2}

    therefor

    2x = \frac{\pi}{2}

    so x = \frac{\pi}{4}

    for  \pi \ge x \ge 0
    Last edited by bigwave; January 21st 2010 at 12:44 PM. Reason: more info
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  3. #3
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    The numerator of cot2x is cos2x.
    This is zero for

    2x=\frac{\pi}{2}\ and\ \frac{3\pi}{2}
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  4. #4
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    Quote Originally Posted by Archie Meade View Post
    The numerator of cot2x is cos2x.
    This is zero for

    2x=\frac{\pi}{2}\ and\ \frac{3\pi}{2}

    Thanks,

    I have  cot2x = \frac{sin2x}{cos2x} = 0

    wouldn't I have to solve for x? so I would have to multiply both sides my cos2x to get  sin2x = 0 , which is zero at  \pi and 2\pi ?

    And so there are four solutions altogether?
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  5. #5
    Super Member bigwave's Avatar
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    cotx = cosx\sinx

    \cot\theta = \frac{cos\theta}{sin\theta}
    Last edited by bigwave; January 21st 2010 at 01:02 PM. Reason: correct equation
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  6. #6
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    Quote Originally Posted by bigwave View Post
    as mentioned by Archie

    cot2x = \frac{1}{cos2x} = 0

    oh right, Can you please explain how did you get cot2x to equal  \frac{1}{cos2x} ?

    Thank you!
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  7. #7
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    That one is the ratio for Tan2x, Tweety, ie \frac{Sin2x}{Cos2x}

    Cot2x=\frac{Cos2x}{Sin2x}

    Cot2x=\frac{1}{Tan2x}=\frac{1}{\frac{Sin2x}{Cos2x}  }

    which is Tan2x inverted.

    So the other two solutions are found by solving Cos2x=0,
    since Cot2x=0 when the numerator Cos2x is zero.
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  8. #8
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    Actually I was just wondering,

    I have  cot2x = 0

    that would be the same as witting  \frac{1}{tan2x} = 0

    ...meaning tan2x couldn't = 0 ? So how come these other to solutions are 'valid'?

    Can someone please explain, thanks.
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  9. #9
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    You may be thinking of Tan2x going to infinity,
    but yes it does when Cos2x goes to zero.

    \frac{1}{x} goes to zero when x goes to infinity.
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  10. #10
    Super Member bigwave's Avatar
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    Quote Originally Posted by Tweety View Post
    oh right, Can you please explain how did you get cot2x to equal  \frac{1}{cos2x} ?

    Thank you!
    sorry this was not correct...

    already answered tho.
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  11. #11
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    Okay, I kind of 'get it' , thanks for your help.
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  12. #12
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    Cot2x=\frac{1}{Tan2x}

    Tan2x=\frac{Sin2x}{Cos2x}

    \frac{1}{\frac{Sin2x}{Cos2x}}=\frac{\frac{Sin2x}{C  os2x}\frac{Cos2x}{Sin2x}}{\frac{Sin2x}{Cos2x}}=\fr  ac{\frac{Sin2x}{Cos2x}}{\frac{Sin2x}{Cos2x}}\frac{  Cos2x}{Sin2x}=\frac{Cos2x}{Sin2x}

    You can think it through with little fractions first.

    \frac{1}{2}=0.5

    \frac{1}{1}=1

    \frac{1}{0.5}=\frac{1}{\frac{1}{2}}=\frac{(2)\frac  {1}{2}}{\frac{1}{2}}=2

    since the numerator is twice the denominator.

    To divide by a fraction, just turn it upside down and multiply instead.

    \frac{\frac{3}{4}}{\frac{2}{5}}=\frac{3}{4}\frac{5  }{2}=\frac{15}{8}

    Therefore

    Cot2x=\frac{1}{Tan2x}

    is Tan2x upside down.
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