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    Find roots of the equation mx=sin x considering different values of m.

    This was an interview question so I don't have a calculator and cant use iteration. For m>1 I found there is only one trivial solution at x=0 (as x>sin x for all x>0 and x<sin x for all x<0). Any suggestions?
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    (Original post by anshul95)
    Find roots of the equation mx=sin x considering different values of m.

    This was an interview question so I don't have a calculator and cant use iteration. For m>1 I found there is only one trivial solution at x=0 (as x>sin x for all x>0 and x<sin x for all x<0). Any suggestions?
    Consider the values of m where there is only one intersection of mx and sinx. Note that the line is tangent to the curve at those points so what can you tell me about the gradients of mx and sinx at those points? Find the value of m for which that is true.
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    draw a graaaaaph
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    (Original post by Farhan.Hanif93)
    Consider the values of m where there is only one intersection of mx and sinx. Note that the line is tangent to the curve at those points so what can you tell me about the gradients of mx and sinx at those points? Find the value of m for which that is true.
    m=cos x so I need to consider the x such that the tangent through sin x at x passes through the origin. I was tempted to do this but that means you would get tan x=x which I can't solve using non-calculator methods.
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    (Original post by anshul95)
    m=cos x so I need to consider the x such that the tangent through sin x at x passes through the origin. I was tempted to do this but that means you would get tan x=x which I can't solve using non-calculator methods.
    At the origin, we have m=\cos (0) = 1. Therefore the line y=x is tangent to y=sinx and we have one intersection. Now what would happen as m gets bigger than one? How many solutions do you have when m is zero? What are those solutions? For what other values of m is the line 'y=mx' a tangent to y=sinx in the range 0&lt;m&lt;1? How many solutions are there in these cases? This question has quite a lot to it.
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    (Original post by Farhan.Hanif93)
    At the origin, we have m=\cos (0) = 1. Therefore the line y=x is tangent to y=sinx and we have one intersection. Now what would happen as m gets bigger than one? How many solutions do you have when m is zero? What are those solutions? For what other values of m is the line 'y=mx' a tangent to y=sinx in the range 0&lt;m&lt;1? How many solutions are there in these cases? This question has quite a lot to it.
    Yes there does seem a lot to this question. I did a similar question about e^x=kx and thought that the same principles apply seems to be a lot more complicated though
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    As someone has allready said, think of the graphs.
    The way I look at this problem is to think of the point at which the the y=mx graph reaches y=1. There will be no more solutions past that x value.
    Thats how I got to an answer.
    I hope that helps though there are other people on this forum much more qualified than me to comment.
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    I would think about gradients of the tangents to  y = \sin x at points where the tangent passes through 0, 0 - the further away from  x = 0 these points are, the smaller the gradient, and they would intersect  y = \sin x more often
    Not sure if this is useful at all though, it's just my thoughts
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    (Original post by dnumberwang)
    I would think about gradients of the tangents to  y = \sin x at points where the tangent passes through 0, 0 - the further away from  x = 0 these points are, the smaller the gradient, and they would intersect  y = \sin x more often
    Not sure if this is useful at all though, it's just my thoughts
    That's the right train of thought.
 
 
 
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