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    I'm doing Paper Q of the IYGB papers and am having trouble with question 8b.

    The paper and mark scheme are here http://www.madasmaths.com/archive/iy...apers/c3_q.pdf and http://www.madasmaths.com/archive/iy..._solutions.pdf

    I get to the stage 2 theta - pi/3 = theta and then I can find the first solution, theta = pi/3. But after this I don't know how to find the other solutions.

    Also, could a question like this be on an actual Edexcel C3 paper?

    Thanks
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    (Original post by anonwinner)
    I'm doing Paper Q of the IYGB papers and am having trouble with question 8b.

    The paper and mark scheme are here http://www.madasmaths.com/archive/iy...apers/c3_q.pdf and http://www.madasmaths.com/archive/iy..._solutions.pdf

    I get to the stage 2 theta - pi/3 = theta and then I can find the first solution, theta = pi/3. But after this I don't know how to find the other solutions.

    Also, could a question like this be on an actual Edexcel C3 paper?

    Thanks
    Do you remember how to solve trig equations of the form \cos x= k?

    We have that x = \arccos k + 2n\pi but also (because fourth quadrant) x = 2\pi - \arccos k + 2\npi.
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    (Original post by Zacken)
    Do you remember how to solve trig equations of the form \cos x= k?

    We have that x = \arccos k + 2n\pi but also (because fourth quadrant) x = 2\pi - \arccos k + 2\npi.
    I use the CAST diagram method.

    After I had found theta = pi/3 I then wrote theta = 2pi - pi/3 to find the solution in the fourth quadrant.

    This resulted in theta = 5pi/3 but this is not a correct solution.
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    (Original post by anonwinner)
    I use the CAST diagram method.

    After I had found theta = pi/3 I then wrote theta = 2pi - pi/3 to find the solution in the fourth quadrant.

    This resulted in theta = 5pi/3 but this is not a correct solution.
    You found 2\theta - \frac{\pi}{3} = \theta (the \theta on your RHS is the "solution").

    So you need to do: 2\theta - \frac{\pi}{3} = 2\pi - \theta now.

    It might help for you to write 2\theta -\frac{\pi}{3} = x to help your visualisation.
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    (Original post by Zacken)
    You found 2\theta - \frac{\pi}{3} = \theta (the \theta on your RHS is the "solution".

    So you need to do: 2\theta - \frac{\pi}{3} = 2\pi - \theta now.

    It might help for you to write 2\theta -\frac{\pi}{3} = x to help your visualisation.
    Thanks I think I am starting to understand.

    If this had been a typical C3 trig question (with theta on only 1 side of the equation), then my method would have worked. Why does it not work for this question?
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    (Original post by anonwinner)
    Thanks I think I am starting to understand.

    If this had been a typical C3 trig question (with theta on only 1 side of the equation), then my method would have worked. Why does it not work for this question?
    Because \theta is on both sides of the equation. You wouldn't get a question like that in an Edexcel exam though, or at least it would be very unlikely.
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    (Original post by Zacken)
    Because \theta is on both sides of the equation. You wouldn't get a question like that in an Edexcel exam though, or at least it would be very unlikely.
    Okay thanks, I really hope it's not on the paper this year
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    (Original post by anonwinner)
    Okay thanks, I really hope it's not on the paper this year
    It won't be. Just remember, all your CAST stuff happens before manipulating the equation, if that makes sense.
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    CAST is what you support a limb after a fracture to aid recovery
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    (Original post by TeeEm)
    CAST is what you support a limb after a fracture to aid recovery
    Top banter m8
 
 
 
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