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    QP:
    http://filestore.aqa.org.uk/subjects...W-QP-JAN11.PDF

    MS:
    http://filestore.aqa.org.uk/subjects...W-MS-JAN11.PDF

    Question 4.

    This is my working out.
    Inverse sin, got -pi/6.

    Therefore.
    In my book it says: x = 2piN + (pi - alpha).

    4x - 2pi/3 = 2piN + pi + pi/6.
    4x = 2piN + 11pi/6.
    x = .5piN + 11pi/24

    So why is this answer not in the mark scheme? Have I made a mistake somewhere?
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    (Original post by Konnichiwa)
    QP:
    http://filestore.aqa.org.uk/subjects...W-QP-JAN11.PDF

    MS:
    http://filestore.aqa.org.uk/subjects...W-MS-JAN11.PDF

    Question 4.

    This is my working out.
    Inverse sin, got -pi/6.

    Therefore.
    In my book it says: x = 2piN + (pi - alpha).

    4x - 2pi/3 = 2piN + pi + pi/6.
    4x = 2piN + 11pi/6.
    x = .5piN + 11pi/24

    So why is this answer not in the mark scheme? Have I made a mistake somewhere?
    Generally speaking you get two possible values of theta in the range from 0 to 2pi when you solve sin(theta) = some value.

    You can add 2pi to each of these values to produce a general solution. There is a way to combine them, but I prefer to answer with two families.
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    I'm not sure what formula you're quoting, but the way I do it is:
    \sin(\frac{-\pi}6) = -\frac12 = \sin(\frac{-\pi}6+2\pi n), so 4x-\frac{2\pi}3 = \frac{-\pi}6+2\pi n and so \frac{\pi}6-\frac{\pi}{24}+\frac{\pi n}2 = x gives pi/8 + pi n/2.

    Also \sin(x) = \sin(\pi-x), so we get \sin(\frac{7\pi}6 + 2\pi n) = -\frac12 = \sin(4x - \frac{2\pi}3), and we can repeat the previous steps.

    (I misread the question the first time…)
 
 
 
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