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Math Help - Solving for x with sin

  1. #1
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    Solving for x with sin

    Hi all,

    Any help here is appreciated, I'm just not sure how to get this problem started.

    Solve for all possible values of x -

    \sqrt{2}\sin x + \sin 2x = 0

    Any points in the right direction would be greatly appreciated.
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  2. #2
    Super Member malaygoel's Avatar
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    Quote Originally Posted by Peleus View Post
    Hi all,

    Any help here is appreciated, I'm just not sure how to get this problem started.

    Solve for all possible values of x -

    \sqrt{2}\sin x + \sin 2x = 0

    Any points in the right direction would be greatly appreciated.
    \sqrt{2}\sin x + \sin 2x = 0

    \sqrt{2}\sin x + 2\sin x \cos x = 0

    \sin x(\sqrt{2} + 2 \cos x) = 0
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  3. #3
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    Just to add a bit of direction here; remember that most trig problems like this are all testing your ability to recall Trigonometric Identities and relationships. In this case, you are meant to use the Double Angle Identity for sine which says:

    sin2\theta=2cos\theta sin\theta
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  4. #4
    Super Member dhiab's Avatar
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    Lightbulb

    Quote Originally Posted by malaygoel View Post
    \sqrt{2}\sin x + \sin 2x = 0

    \sqrt{2}\sin x + 2\sin x \cos x = 0

    \sin x(\sqrt{2} + 2 \cos x) = 0
    Hello : Thank you ...
    here is the suite of resolution :


    \sin x\left( {\sqrt 2 + 2\cos x} \right) = 0 \Leftrightarrow \left( {\sin x = 0} \right) \vee \left( {\sqrt 2 + 2\cos x = 0} \right)
    \sin x = 0 \Leftrightarrow x = n\pi ....(n \in {\rm Z})

     <br />
\begin{array}{l}<br />
\sqrt 2 + 2\cos x = 0 \Leftrightarrow \cos x = - \frac{{\sqrt 2 }}{2} \\ <br />
\cos x = - \frac{{\sqrt 2 }}{2} \Leftrightarrow \cos x = - \cos \frac{\pi }{4} \\ <br />
\end{array}<br />


    \begin{array}{l}<br />
\cos x = - \frac{{\sqrt 2 }}{2} \Leftrightarrow \cos x = \cos \left( {\pi - \frac{\pi }{4}} \right) \\ <br />
\cos x = - \frac{{\sqrt 2 }}{2} \Leftrightarrow \left( {x = \frac{{3\pi }}{4} + 2n\pi } \right)or\left( {x = - \frac{{3\pi }}{4} + 2n\pi } \right)....(n \in {\rm Z}) \\ <br />
\end{array}
    Thanks
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  5. #5
    Senior Member pacman's Avatar
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    the graph looks like this, can anyone interpret this? are the answers above represents the graph?
    Attached Thumbnails Attached Thumbnails Solving for x with sin-ha.gif  
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  6. #6
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    Quote Originally Posted by pacman View Post
    the graph looks like this, can anyone interpret this? are the answers above represents the graph?
    that's not the graph of y = \sqrt{2} \sin{x} + \sin(2x)
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  7. #7
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    Quote Originally Posted by skeeter View Post
    that's not the graph of y = \sqrt{2} \sin{x} + \sin(2x)
    Both graphs are right. You just have a better "window".
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  8. #8
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    Quote Originally Posted by ANDS! View Post
    Both graphs are right. You just have a better "window".
    you're right ... need to look closer next time.
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  9. #9
    Super Member dhiab's Avatar
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    Quote Originally Posted by pacman View Post
    the graph looks like this, can anyone interpret this? are the answers above represents the graph?
    Hello : that's the graph of
    Attached Thumbnails Attached Thumbnails Solving for x with sin-1616.jpg  
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  10. #10
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    Quote Originally Posted by dhiab View Post
    Hello : that's the graph of
    and now we have three.
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  11. #11
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    Quote Originally Posted by skeeter View Post
    that's not the graph of y = \sqrt{2} \sin{x} + \sin(2x)

    what software did you guys use to generate these graphs ?
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  12. #12
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