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Math Help - Tricky sin equation

  1. #1
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    Smile Tricky sin equation

    Hello!
    This is the equation Im stuck with:
    2sin (0.5x) =  3/3x+2


    Im not sure where to start, because I cant take out any of the x values,

    should i let 0.5x be a completely new vale like 'y' and but then i will still be stuck


    Thank you!
    Last edited by mr fantastic; April 23rd 2010 at 02:47 PM. Reason: Edited post title
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  2. #2
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    Quote Originally Posted by appleseed View Post
    Hello!
    This is the equation Im stuck with:
    2sin (0.5x) =  3/3x+2


    Im not sure where to start, because I cant take out any of the x values,

    should i let 0.5x be a completely new vale like 'y' and but then i will still be stuck


    Thank you!
    What is required in the problem? Do you want to find the value of x? If yes, how much decimal places in x is required ?
    In the form of infinite series sinθ can be written as
    sinθ = θ - (θ)^3/3! + (θ)^5/5! and so on.
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  3. #3
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    Quote Originally Posted by sa-ri-ga-ma View Post
    What is required in the problem? Do you want to find the value of x? If yes, how much decimal places in x is required ?
    In the form of infinite series sinθ can be written as
    sinθ = θ - (θ)^3/3! + (θ)^5/5! and so on.


    Hello!

    Thank you so much for the reply!!!

    Yes i am asked to find the value of x and the answer is required to 3 s.f.

    It might be better if i give you the context of the equation.


    ''''''''''''There are two equations g(x) and f(x), There are two values of x for which the gradient of f is equal to the gradient
    of g. Find both these values of x. '''''''''"




    So i've already found the derivative of the two equations and i've checked that its correct, and then i equated the two derivatives, but i dont know how to find x.

    I was thinking that maybe it could be done by trig identities, lol but i dont know how though


    anyways Thank you! !!
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  4. #4
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    Hello appleseed

    There are infinitely many solutions to the equation
    2\sin(\tfrac12x)=\frac{3}{3x+2}
    which is what I assume you meant. (You should have written 3/(3x+2) if you don't know how to write it using LaTeX.)

    Have a look at the diagram I've attached, where I've plotted the graph of each side separately.


    You won't be able to solve an equation like this to get exact answers. You'll have to use a numerical method - e.g. Newton-Raphson. Do you know how?


    Grandad
    Attached Thumbnails Attached Thumbnails Tricky sin equation-untitled.jpg  
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  5. #5
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    Quote Originally Posted by Grandad View Post
    Hello appleseed

    There are infinitely many solutions to the equation
    2\sin(\tfrac12x)=\frac{3}{3x+2}
    which is what I assume you meant. (You should have written 3/(3x+2) if you don't know how to write it using LaTeX.)

    Have a look at the diagram I've attached, where I've plotted the graph of each side separately.

    You won't be able to solve an equation like this to get exact answers. You'll have to use a numerical method - e.g. Newton-Raphson. Do you know how?

    Grandad


    Hi Grandad!!!
    Thank you for the reply.

    I think i've heard of Newton-Raphson before, but i cant remember exactly how to find the answer using it. Could you briefly go over the steps please?


    Thanks!
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  6. #6
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    Newton-Raphson method

    Hello appleseed
    Quote Originally Posted by appleseed View Post
    Hi Grandad!!!
    Thank you for the reply.

    I think i've heard of Newton-Raphson before, but i cant remember exactly how to find the answer using it. Could you briefly go over the steps please?


    Thanks!
    If you've not had any practice with this method, this is not the easiest example to start with.

    If you want to get a solution to the equation
    f(x) = 0
    and you have a first approximation, x = a_1, to the answer, then a second approximation, a_2, is given by
    a_2=a_1-\frac{f(a_1)}{f'(a_1)}
    In this case you'll have to let
    f(x) = 2\sin(\tfrac12x)-\frac{3}{3x+2}
    So
    f'(x) = \cos(\tfrac12x) +\frac{9}{(3x+2)^2}
    From the graph, there's a solution close to x = 1. So with a_1 = 1 we get
    a_2 = 1-\frac{f(1)}{f'(1)}
    \approx 0.7
    ... and so on.

    The next approximation, a_3, uses 0.7 as the starting value:
    a_3 = 0.7 -\frac{f(0.7)}{f'(0.7)}
    \approx 0.73 (to 2 d.p.)
    Repeating the process, to 4 d.p., I make the answer 0.7314.

    The next positive solution is around x = 6. To 4 d.p., the solution converges very quickly to 6.1361.

    Grandad
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