Results 1 to 12 of 12

Thread: trig equation - all solutions

  1. #1
    Super Member
    Joined
    Sep 2008
    Posts
    631

    trig equation - all solutions

    solve $\displaystyle cosec^{2}2x - cot2x = 1 $

    $\displaystyle 0 \leq x \leq 180 $


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

    $\displaystyle 1 + cot^{2}2x -cot2x = 1 $

    $\displaystyle cot^{2}2x -cot2x = 0 $

    $\displaystyle cot2x(cot2x-1) = 0 $

    $\displaystyle cot2x = 0 $ or

    $\displaystyle cot2x = 1 $

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

    for the other one I got $\displaystyle tan2x = 1 $

    $\displaystyle 2x = 45, 225 $

    $\displaystyle x = 22.5 , 112.5 $

    Are these all the solutions in the interval, or does cot2x =0 also have a solution?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member bigwave's Avatar
    Joined
    Nov 2009
    From
    Wahiawa, Hawaii
    Posts
    679
    Thanks
    1
    Quote Originally Posted by Tweety View Post
    solve $\displaystyle cosec^{2}2x - cot2x = 1 $

    $\displaystyle 0 \leq x \leq 180 $


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

    $\displaystyle 1 + cot^{2}2x -cot2x = 1 $

    $\displaystyle cot^{2}2x -cot2x = 0 $

    $\displaystyle cot2x(cot2x-1) = 0 $

    $\displaystyle cot2x = 0 $ or

    $\displaystyle cot2x = 1 $

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

    for the other one I got $\displaystyle tan2x = 1 $

    $\displaystyle 2x = 45, 225 $

    $\displaystyle x = 22.5 , 112.5 $

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

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

    therefor

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

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

    for$\displaystyle \pi \ge x \ge 0$
    Last edited by bigwave; Jan 21st 2010 at 12:44 PM. Reason: more info
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    4
    The numerator of cot2x is cos2x.
    This is zero for

    $\displaystyle 2x=\frac{\pi}{2}\ and\ \frac{3\pi}{2}$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Sep 2008
    Posts
    631
    Quote Originally Posted by Archie Meade View Post
    The numerator of cot2x is cos2x.
    This is zero for

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

    Thanks,

    I have $\displaystyle 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 $\displaystyle sin2x = 0 $ , which is zero at $\displaystyle \pi $ and $\displaystyle 2\pi $?

    And so there are four solutions altogether?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member bigwave's Avatar
    Joined
    Nov 2009
    From
    Wahiawa, Hawaii
    Posts
    679
    Thanks
    1

    cotx = cosx\sinx

    $\displaystyle \cot\theta = \frac{cos\theta}{sin\theta}$
    Last edited by bigwave; Jan 21st 2010 at 01:02 PM. Reason: correct equation
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Sep 2008
    Posts
    631
    Quote Originally Posted by bigwave View Post
    as mentioned by Archie

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

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

    Thank you!
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    4
    That one is the ratio for Tan2x, Tweety, ie $\displaystyle \frac{Sin2x}{Cos2x}$

    $\displaystyle Cot2x=\frac{Cos2x}{Sin2x}$

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

    which is Tan2x inverted.

    So the other two solutions are found by solving Cos2x=0,
    since Cot2x=0 when the numerator Cos2x is zero.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member
    Joined
    Sep 2008
    Posts
    631
    Actually I was just wondering,

    I have $\displaystyle cot2x = 0 $

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

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

    Can someone please explain, thanks.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    4
    You may be thinking of Tan2x going to infinity,
    but yes it does when Cos2x goes to zero.

    $\displaystyle \frac{1}{x}$ goes to zero when x goes to infinity.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member bigwave's Avatar
    Joined
    Nov 2009
    From
    Wahiawa, Hawaii
    Posts
    679
    Thanks
    1
    Quote Originally Posted by Tweety View Post
    oh right, Can you please explain how did you get cot2x to equal $\displaystyle \frac{1}{cos2x} $ ?

    Thank you!
    sorry this was not correct...

    already answered tho.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Super Member
    Joined
    Sep 2008
    Posts
    631
    Okay, I kind of 'get it' , thanks for your help.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    4
    $\displaystyle Cot2x=\frac{1}{Tan2x}$

    $\displaystyle Tan2x=\frac{Sin2x}{Cos2x}$

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

    You can think it through with little fractions first.

    $\displaystyle \frac{1}{2}=0.5$

    $\displaystyle \frac{1}{1}=1$

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

    since the numerator is twice the denominator.

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

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

    Therefore

    $\displaystyle Cot2x=\frac{1}{Tan2x}$

    is Tan2x upside down.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. All solutions for trig equation
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: Sep 13th 2011, 05:52 AM
  2. Finding all solutions to the trig equation
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: Dec 11th 2010, 10:00 PM
  3. Trig equation (and its solutions)
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: Sep 25th 2008, 08:54 AM
  4. evaluating solutions to trig equation??
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: Aug 22nd 2008, 07:41 AM
  5. Trig equation - graphing to show solutions
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: May 1st 2008, 08:31 AM

Search Tags


/mathhelpforum @mathhelpforum