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Math Help - find solution qn

  1. #1
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    find solution qn

    Given that \frac{\pi}{10} is 1 of the solutions for the eqn sin3x=cos2x for 0<x<\pi, find the 2 other solutions. Leave ur ans in terms of \pi.

    thanks for reading.
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  2. #2
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    Remember that \sin{3x} = 3\sin{x} - 4\sin^3{x} and \cos{2x} = 1 - 2\sin^2{x}.

    Then the equation

    \sin{3x} = \cos{2x}

    is the same as

    3\sin{x} - 4\sin^3{x} = 1 - 2\sin^2{x}

    0 = 4\sin^3{x} - 2\sin^2{x} - 3\sin{x} + 1.


    This is a cubic in \sin{x}. So let X = \sin{x} to give

    0 = 4X^3 - 2X^2 - 3X + 1.

    It's easy to see that X = 1 satisfies this equation.

    So \sin{x} = 1 meaning x = \frac{\pi}{2} is a solution.


    You are also told that \frac{\pi}{10} is a solution. This is in the first quadrant. But your domain also includes the second quadrant.

    So \pi - \frac{\pi}{10} = \frac{9\pi}{10} is also a solution.


    So your solutions are x = \left\{\frac{\pi}{10}, \frac{\pi}{2}, \frac{9\pi}{10}\right\}.
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  3. #3
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    \sin{3x} = 3\sin{x} - 4\sin^3{x}
    what formula is that?


    You are also told that \frac{\pi}{10} is a solution. This is in the first quadrant. But your domain also includes the second quadrant.

    So \pi - \frac{\pi}{10} = \frac{9\pi}{10} is also a solution.
    Dun understand this part seriously.
    Last edited by mr fantastic; September 12th 2010 at 04:21 AM. Reason: Merged posts.
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  4. #4
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    I think you know that

    sin(A+B) = sinAcosB + cosAsinB

    If A + B = x, then sin(2x) = 2sin(x)cos(x)

    If you write sin(3x) as sin(2x+x) and expand and simplify, you will get above formula.
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  5. #5
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    I dun get ur point either.

    Prove that  sin{3x} = 3\sin{x} - 4\sin^3{x}

    sin(2x+x)=sin(2x)cos(x)+cos(2x)sin(x)
    =2sin(x)cos(x)cos(x)+cos(2x)sin(x)
    =2sin(x)cos^2(x)+cos(2x)sin(x)
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  6. #6
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    Quote Originally Posted by stupidguy View Post
    Dun understand this part seriously.
    You can have a different method.

    cos(2x) = sin(3x) = cos(π/2 - 3x) Hence the general solution will be

    2x = 2nπ + π/2 - 3χ or 2x = 2nπ -π/2 + 3x. The second one gives the negative x. So leave it.

    5x = 2nπ + π/2. Put n = 0, 1 and 2 and find the values of x.
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  7. #7
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    Quote Originally Posted by stupidguy View Post
    I dun get ur point either.

    Prove that  sin{3x} = 3\sin{x} - 4\sin^3{x}

    sin(2x+x)=sin(2x)cos(x)+cos(2x)sin(x)
    =2sin(x)cos(x)cos(x)+cos(2x)sin(x)
    =2sin(x)cos^2(x)+cos(2x)sin(x)
    = 2sin(x)[1-sin^2(x)] + [1 -2sin^2(x)]sin(x)

    = 2sin(x) - 2sin^3(x) + sin(x) - 2sin^3(x)

    = 3sin(x) - 4sin^3(x)
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  8. #8
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    Quote Originally Posted by sa-ri-ga-ma View Post
    You can have a different method.

    cos(2x) = sin(3x) = cos(π/2 - 3x) Hence the general solution will be

    2x = 2nπ + π/2 - 3χ or 2x = 2nπ -π/2 + 3x. The second one gives the negative .o leave it.

    5x = 2nπ + π/2. Put n = 0, 1 and 2 and find the values of x.
    I was thinking of sketching 2 graphs and finding the intersections.

    The first intersection is \dfrac{\pi}{10} (given), the second is 0.5 (obvious).

    My real doubt is how to get the last solution which is not readable from the sketch.

    According to Prove It,

    \pi - \frac{\pi}{10} = \frac{9\pi}{10}

    Why?
    Last edited by mr fantastic; September 12th 2010 at 04:23 AM. Reason: Background that might guide the type of help given should be put in the first post.
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  9. #9
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    you guys actually predicted and solved the qn part 2 b4 i asked which is

    show that sin3x=3sinx-4sin^3{x} and hence prove that

    cos(2x)-sin(3x)\equiv 4\sin^3{x} - 2\sin^2{x} - 3\sin{x} + 1

    i believe that part 1 which is the original qn do not require part 2 working and is also rather simple cos it is only 2 marks, though I dunno how to do.
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  10. #10
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    OK. Now you know that

    If sin(x) = 1/2, then x can be either π/6 or (π - π/6) = 5π/6

    Similarly if x = π/10 satisfies the given equation, then x = (π-π/10) = 9π/10 must also satisfy the given equation.
    Last edited by sa-ri-ga-ma; September 11th 2010 at 03:44 AM.
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  11. #11
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    You should know that \sin{\theta} is positive in the first and second quadrants, and that by symmetry, \sin{\left(\pi - \theta\right)} = \sin{\theta}.

    In this case, \theta = \frac{\pi}{10}.
    Last edited by mr fantastic; September 12th 2010 at 04:26 AM.
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  12. #12
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    these are my "sketches".

    find solution qn-untitled.jpg

    I know that A=pi/10, how do I know B=9pi/10 assuming I sketch?

    In the first place, I would not know that they are of the same y-value, hence I cannot determine their x-value like what you done.
    Last edited by stupidguy; September 11th 2010 at 04:00 AM.
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  13. #13
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    You can't read from your sketch that the two graphs cross at x = 0.9\pi = \frac{9\pi}{10}?


    Look at it this way. You have a domain of 0 < x< \pi. This means that your solutions are in the first quadrant. So any sine values are positive.

    Going back to the equation that we found using the identities I listed in my first post...

    0 = 4X^3 - 2X^2 - 3X + 1

    0 = (X - 1)(4X^2 + 2X - 1).


    That means 4X^2 + 2X - 1 = 0.

    This gives X = \frac{-1 \pm \sqrt{5}}{4} after applying the Quadratic formula.

    So \sin{x} = \frac{-1 \pm \sqrt{5}}{4} since X = \sin{x}.


    Now you need to remember from your domain of 0 < x < \pi that \sin{x} is positive.

    So we can only accept

    \sin{x} = \frac{-1 + \sqrt{5}}{4}.

    The first solution x = \frac{\pi}{10} is in the first quadrant.

    You also have another solution in the second quadrant. How would you find it? Think about a unit circle.


    Edit: In fact, here is a picture of a unit circle, with the angle of x = \frac{\pi}{10}. Can you see that there is another value in the second quadrant where the sine value (the vertical lengths in red) is the same? What is that angle (measured from the positive x axis)?

    Last edited by Prove It; September 11th 2010 at 04:29 AM.
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  14. #14
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    Nice sketches.

    From the sketches you can clearly see that at A and B, y values are the same for both the graphs. Find the corresponding values of x.

    For C, y value is different but it is same for both the graphs.
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  15. #15
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    Notice I added inverted commas for sketches because in the exams we do not use graphic calculator.

    So assuming I sketched the 2 graphs, point A and B will not be accurate (not having the same y value).

    I cant read those values off my actual sketch so how do i reason it?
    Last edited by stupidguy; September 11th 2010 at 05:14 AM.
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