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Math Help - More trig review

  1. #1
    Super Member 11rdc11's Avatar
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    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) 3\sin^{x} - \sin{x} - 1 =0

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

    C) 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

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

    so

    \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; November 9th 2008 at 08:51 PM.
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  2. #2
    Super Member flyingsquirrel's Avatar
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    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) 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 :

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

    <br />
2\sin x -3\cos x = \sqrt{13}\left(\sin x\cos \phi -\sin\phi\cos x\right)<br />
=\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.
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  3. #3
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by flyingsquirrel View Post
    Hi,

    I think it is correct too.



    I'd rather do it this way :


    \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}
    " alt="
    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 \left(\tfrac{2}{\sqrt{13}}\right)^2+\left(\tfrac{3  }{\sqrt{13}}\right)^2=1 so there exists \phi\in[0,2\pi[ such that \begin{cases}\sin \phi = \tfrac{3}{\sqrt{13}}\\ \cos \phi=\tfrac{2}{\sqrt{13}} \end{cases}. This gives us

    2\sin x -3\cos x = \sqrt{13}\left(\sin x\cos \phi -\sin\phi\cos x\right)
    =\sqrt{13}\sin(x-\phi)
    " alt="
    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.
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  • #4
    Super Member flyingsquirrel's Avatar
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    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, 3.84=-2.43+2\pi and 5.02=-1.26+2\pi because \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 (2\sin x)^2=(1+3\cos x)^2 that is to say 2\sin x =\pm(1+3\cos x). However we can easily find the solutions of 2\sin x =+(1+3\cos x) : these are the solutions of 2\sin x =\pm(1+3\cos x) such that 2\sin x and 1+3\cos x share the same sign...
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