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    Very easy question however I don't get the right answer. I've asked my friends who also do FP1 and they get the same as me.

    Question 4: http://filestore.aqa.org.uk/subjects...W-QP-JAN11.PDF

    I got

    x= π/8 + (πn)/2
    x= (11π)/24 + (πn)/2

    However the mark scheme suggests:

    x= π/8 + (πn)/2
    x= (-π)/24 + (πn)/2
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    QUOTE=2014_GCSE;65789279]Very easy question however I don't get the right answer. I've asked my friends who also do FP1 and they get the same as me.

    Question 4: http://filestore.aqa.org.uk/subjects...W-QP-JAN11.PDF

    I got

    x= π/8 + (πn)/2
    x= (11π)/24 + (πn)/2

    However the mark scheme suggests:

    x= π/8 + (πn)/2
    x= (-π)/24 + (πn)/2[ QUOTE]

    Just tried it and got the solution from the markscheme - want me to send over a pic? :P
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    (Original post by Chickenslayer69)
    QUOTE=2014_GCSE;65789279]Very easy question however I don't get the right answer. I've asked my friends who also do FP1 and they get the same as me.

    Question 4: http://filestore.aqa.org.uk/subjects...W-QP-JAN11.PDF

    I got

    x= π/8 + (πn)/2
    x= (11π)/24 + (πn)/2

    However the mark scheme suggests:

    x= π/8 + (πn)/2
    x= (-π)/24 + (πn)/2[ QUOTE]

    Just tried it and got the solution from the markscheme - want me to send over a pic? :P
    YES PLEASE! Basically, I got P.V as -1/6pi and so you do 2pi(n) - 1/6pi and then 2pi(n) + pi - -1/6pi which gives you 2pi(n) + 7/6pi.
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    if you dont understand something ive done just ask sry for handwriting lool
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    (Original post by Chickenslayer69)
    Name:  kokk.png
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    if you dont understand something ive done just ask sry for handwriting lool
    We always learned it, that you get the two values by the the initial value (in this case -1/6pi) and then 2pi - the initial value (which in this case would be 7/6pi). So why doesn't this work? Is it because you have to do it within a certain domain?
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    QUOTE=2014_GCSE;65790239]We always learned it, that you get the two values by the the initial value (in this case -1/6pi) and then 2pi - the initial value (which in this case would be 7/6pi). So why doesn't this work? Is it because you have to do it within a certain domain?[ QUOTE]

    Drawing the sin graph in this case works - you can see from the graph I drew that the initial value (-1/2) occurs at -pi + 1/6pi = -5/6pi, and the other value is 1/6pi as you have already worked out. 7/6pi would be impossible since it is out of the pi range either way (values for general solution can only be from -pi and pi unless you are given an interval). So you always look for the values the initial value (in this case -1/2) gives in the range from -pi to pi.

    Hope this helps
 
 
 
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