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    For Q3b.ii., can someone explain to me why 90* is an answer... (I had 45* and 135*)
    QP:
    http://filestore.aqa.org.uk/subjects...4-QP-JAN13.PDF
    MS:
    http://filestore.aqa.org.uk/subjects...W-MS-JAN13.PDF
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    (Original post by pippabethan)
    For Q3b.ii., can someone explain to me why 90* is an answer... (I had 45* and 135*)
    QP:
    http://filestore.aqa.org.uk/subjects...4-QP-JAN13.PDF
    MS:
    http://filestore.aqa.org.uk/subjects...W-MS-JAN13.PDF
    When cotx = 0.
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    If you divided the equation by cosx (once you converted sin2x into 2sinxcosx) you shouldn't have.
    I'm assuming that's what you did.
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    (Original post by SeanFM)
    When cotx = 0.
    But tanx=1/0 isn't possible?
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    (Original post by pippabethan)
    But tanx=1/0 isn't possible?
    If you write cotx in terms of its components, then..
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    (Original post by SeanFM)
    If you write cotx in terms of its components, then..
    Oh I see, so you write cotx as cosx over sinx
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    (Original post by pippabethan)
    Oh I see, so you write cotx as cosx over sinx
    Correct. :borat:
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    basically yh

    cos x
    ------- - 2sinx = 0
    sinx

    then u write sin2x in terms of sinx and cosx using double angle formula

    then u should be able to solve by factoring, you would get 0 and 90* as shown in the markscheme but you discard 0 as its not in the range
 
 
 
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