Results 1 to 8 of 8

Thread: general solution ,trig?

  1. #1
    Junior Member
    Joined
    Oct 2008
    Posts
    74

    general solution ,trig?

    What is the general solution to: $\displaystyle 2sin2\Theta=\sqrt{3}tan2\Theta$? any ideas?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member Spec's Avatar
    Joined
    Aug 2007
    Posts
    318
    $\displaystyle 2\sin 2\theta = \sqrt{3}\tan 2\theta = \sqrt{3}\frac{\sin 2\theta}{\cos 2\theta}\Longleftrightarrow \sin 2\theta = 0$ or $\displaystyle \cos 2\theta = \frac{\sqrt{3}}{2}$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Amer's Avatar
    Joined
    May 2009
    From
    Jordan
    Posts
    1,093
    ok

    $\displaystyle 2sin2\theta=\sqrt{3}tan2\theta$


    $\displaystyle 2sin2\theta=\sqrt{3}\left(\frac{sin2\theta}{cos2\t heta}\right)$

    dived the to side sin2 theta

    $\displaystyle 2=\sqrt{3}\left(\frac{1}{cos2\theta}\right)$

    $\displaystyle cos2\theta=\frac{\sqrt{3}}{2}$

    $\displaystyle 2\theta=\frac{\pi}{6}....\theta=\frac{\pi}{12}+2n\ pi $
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Oct 2008
    Posts
    74
    would the end result be $\displaystyle \frac{\pi}{12}+n\pi$ or would it be $\displaystyle \frac{\pi}{12}+2n\pi$?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member Spec's Avatar
    Joined
    Aug 2007
    Posts
    318
    You can't just divide away $\displaystyle \sin 2\theta$ from both sides since it's not a constant. If you do that, you'll miss out on all the solutions where $\displaystyle \sin 2\theta=0$

    It's similar to this equation: $\displaystyle x^3=x$ has obviously got one root $\displaystyle x=0$, but if you divide both sides with $\displaystyle x$ without taking that into consideration you'll end up with $\displaystyle x^2=1$, and thus miss a root.

    The general solution to $\displaystyle \sin x = C$ is $\displaystyle x=v+2n\pi$ or $\displaystyle x=\pi-v+2n\pi$ where $\displaystyle v$ is the angle and $\displaystyle n\in Z$

    The general solution to $\displaystyle \cos x = C$ is $\displaystyle x=\pm v+2n\pi$ where $\displaystyle v$ is the angle and $\displaystyle n\in Z$
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member dhiab's Avatar
    Joined
    May 2009
    From
    ALGERIA
    Posts
    582
    Thanks
    3

    resolution

    good night, I offer you a resolution
    2shared - download R1.pdf
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member dhiab's Avatar
    Joined
    May 2009
    From
    ALGERIA
    Posts
    582
    Thanks
    3
    Quote Originally Posted by Amer View Post
    ok

    $\displaystyle 2sin2\theta=\sqrt{3}tan2\theta$


    $\displaystyle 2sin2\theta=\sqrt{3}\left(\frac{sin2\theta}{cos2\t heta}\right)$

    dived the to side sin2 theta

    $\displaystyle 2=\sqrt{3}\left(\frac{1}{cos2\theta}\right)$

    $\displaystyle cos2\theta=\frac{\sqrt{3}}{2}$

    $\displaystyle 2\theta=\frac{\pi}{6}....\theta=\frac{\pi}{12}+2n\ pi $
    good night, you not to pouver not to divide on sin 2theta
    because sin2 theta can etre egal in zero
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor Amer's Avatar
    Joined
    May 2009
    From
    Jordan
    Posts
    1,093
    I am sorry I should be careful when I answer questions
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. general solution of trig equation
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: Sep 14th 2011, 07:40 AM
  2. Finding the general solution from a given particular solution.
    Posted in the Differential Equations Forum
    Replies: 5
    Last Post: Oct 7th 2009, 01:44 AM
  3. general solution to trig problem
    Posted in the Trigonometry Forum
    Replies: 10
    Last Post: May 18th 2009, 02:28 PM
  4. Replies: 2
    Last Post: May 18th 2009, 12:51 PM
  5. Trig. Finding the general solution
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: Dec 17th 2008, 03:36 PM

Search Tags


/mathhelpforum @mathhelpforum