Results 1 to 4 of 4

Thread: More trig review

  1. #1
    Super Member 11rdc11's Avatar
    Joined
    Jul 2007
    From
    New Orleans
    Posts
    894
    Thanks
    1

    More trig review

    Ok I'm having problems with these 3 problems

    Solve each equation in the interval {0,2pi) and round your answer to the nearest hundredth of a radian

    A) $\displaystyle 3\sin^{x} - \sin{x} - 1 =0$

    B) $\displaystyle 2\sin{x} - 3\cos{x} = 1$

    C) $\displaystyle 3\sin{x} - 5 = 0$

    Ok so for problem A I did this

    I used u of sub and solved the quad equation which gave me

    $\displaystyle \frac{1 \pm \sqrt{13}}{6} = u$

    so

    $\displaystyle \sin{x} = \frac{1 \pm \sqrt{13}}{6}$

    Now I take arcsin of both sides and get

    3.59, .88, 5.83, 2.26

    I believe this is correct



    For problem B I'm not sure where to start.

    I'm guessing either square both sides or equal the right side to cos^2x + sin^2x =1



    For problem C

    I get that it is false due to range

    Any help would be appreciated and thanks in advance.
    Last edited by 11rdc11; Nov 9th 2008 at 08:51 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
    Posts
    802
    Hi,
    Quote Originally Posted by 11rdc11 View Post
    Ok so for problem A I did this
    (...) I believe this is correct
    I think it is correct too.

    Quote Originally Posted by 11rdc11 View Post
    C) $\displaystyle 3\sin{x} - 5 = 0$
    Quote Originally Posted by 11rdc11 View Post
    For problem B I'm not sure where to start.

    I'm guessing either square both sides or equal the right side to cos^2x + sin^2x =1
    I'd rather do it this way :

    $\displaystyle \begin{aligned}
    2\sin x -3\cos x &= \sqrt{2^2+3^2}\left(\frac{2}{\sqrt{2^2+3^2}}\sin x -\frac{3}{\sqrt{2^2+3^2}}\cos x\right)\\
    &=\sqrt{13}\left(\frac{2}{\sqrt{13}}\sin x -\frac{3}{\sqrt{13}}\cos x\right)
    \end{aligned}$
    This is interesting because $\displaystyle \left(\tfrac{2}{\sqrt{13}}\right)^2+\left(\tfrac{3 }{\sqrt{13}}\right)^2=1$ so there exists $\displaystyle \phi\in[0,2\pi[$ such that $\displaystyle \begin{cases}\sin \phi = \tfrac{3}{\sqrt{13}}\\ \cos \phi=\tfrac{2}{\sqrt{13}} \end{cases}$. This gives us

    $\displaystyle
    2\sin x -3\cos x = \sqrt{13}\left(\sin x\cos \phi -\sin\phi\cos x\right)
    =\sqrt{13}\sin(x-\phi)$

    Can you take it from here ?
    Quote Originally Posted by 11rdc11 View Post
    For problem C

    I get that it is false due to range
    Yes, this equation has no solution.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member 11rdc11's Avatar
    Joined
    Jul 2007
    From
    New Orleans
    Posts
    894
    Thanks
    1
    Quote Originally Posted by flyingsquirrel View Post
    Hi,

    I think it is correct too.



    I'd rather do it this way :


    $\displaystyle \begin{aligned}$
    $\displaystyle
    2\sin x -3\cos x &= \sqrt{2^2+3^2}\left(\frac{2}{\sqrt{2^2+3^2}}\sin x -\frac{3}{\sqrt{2^2+3^2}}\cos x\right)\\
    &=\sqrt{13}\left(\frac{2}{\sqrt{13}}\sin x -\frac{3}{\sqrt{13}}\cos x\right)
    \end{aligned}
    $
    This is interesting because $\displaystyle \left(\tfrac{2}{\sqrt{13}}\right)^2+\left(\tfrac{3 }{\sqrt{13}}\right)^2=1$ so there exists $\displaystyle \phi\in[0,2\pi[$ such that $\displaystyle \begin{cases}\sin \phi = \tfrac{3}{\sqrt{13}}\\ \cos \phi=\tfrac{2}{\sqrt{13}} \end{cases}$. This gives us

    $\displaystyle
    2\sin x -3\cos x = \sqrt{13}\left(\sin x\cos \phi -\sin\phi\cos x\right)
    =\sqrt{13}\sin(x-\phi)
    $

    Can you take it from here ?

    Yes, this equation has no solution.
    Thanks but I don't think my lil cousin will understand that method when I try to explain it to him. I approached it like this



    square both sides





    Then I go ahead and use u of substitation for the cosx and get



    and I get 2.43 and 1.26. Then when I check it graphically it tells me my only solution is 1.26 and 3.84. Where am I going wrong and once again thanks in advance.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
    Posts
    802
    and I get 2.43 and 1.26. Then when I check it graphically it tells me my only solution is 1.26 and 3.84. Where am I going wrong and once again thanks in advance.
    In fact there are four solutions : 2.43, 1.26, $\displaystyle 3.84=-2.43+2\pi$ and $\displaystyle 5.02=-1.26+2\pi$ because $\displaystyle \cos x = \alpha \Longleftrightarrow x\equiv\pm\arccos \alpha\pmod{2\pi}$.

    You're probably wondering why we have found four solutions whereas graphically you only see two of them. The reason is that we have solved $\displaystyle (2\sin x)^2=(1+3\cos x)^2$ that is to say $\displaystyle 2\sin x =\pm(1+3\cos x)$. However we can easily find the solutions of $\displaystyle 2\sin x =+(1+3\cos x)$ : these are the solutions of $\displaystyle 2\sin x =\pm(1+3\cos x)$ such that $\displaystyle 2\sin x$ and $\displaystyle 1+3\cos x$ share the same sign...
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Pre-Exam Review: Trig is killing me
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: Jun 20th 2010, 10:11 PM
  2. Trig Review Help
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: Aug 31st 2009, 11:12 PM
  3. Need help on Review
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Apr 5th 2009, 10:53 PM
  4. Trig Review - Clueless
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: Aug 5th 2008, 07:58 PM
  5. Trig Review
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: Apr 4th 2006, 06:38 PM

Search Tags


/mathhelpforum @mathhelpforum