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    • Thread Starter
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    Can someone post a method for working this out?

    I havn't finished my core2 course yet but doing a past paper...

    Perhaps I am missing some crucial rule or something?

    SAFL x
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    Try converting the cos2 x term into a sin2 x term then treating it as a quadratic
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    So...

    2(1 - sin^2(x)) + 1 = 5sin(x)

    and hence

    2sin^2(x) + 5sin(x) - 3 = 0

    Be correct as a starting point then?

    The question wants the solutions between the interval 0 < x < 2pi

    I guess i'll have to wait till we cover this more in class -.-
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    Yeah that looks good, now you can factorise that and solve for x
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    Sweet, my free just finished so i'll try it when i'm home x
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    You may find it easier to substitute something like y=sin(x) to make is a bit easier to look at. Then when you have your two values for y, you can make those equal to sin(x).
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    why does the sin^2 get ignored I don;t understand
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    (Original post by sofiane567)
    why does the sin^2 get ignored I don;t understand
    2cos2(x) + 1 = 5sin(x)

    2 [1 - sin2(x)] + 1 = 5sin(x)

    2 - 2sin2(x) + 1 = 5sin(x)

    2sin2(x) + 5sin(x) - 3 = 0

    Let y = sin (x).

    2y2 + 5y - 3 = 0

    (2y-1)(y+3) = 0

    y = \frac{1}{2} or y = -3

    y = sin(x),

    therefore, sin (x) = \frac{1}{2} or sin (x) = -3

    sin-1 (\frac{1}{2}) = 30

    sin-1 (-3) = Maths error.

    0 < x < 2 pi

    x = 30, 150 degrees

    Convert 30, 150 into radians.

    2 pi = 360 degrees

    pi = 180 degrees

    \frac{pi}{6} = 30 degrees

    \frac{5 pi}{6} = 150 degrees.

    Hence, x = \frac{pi}{6} and \frac{5 pi}{6}
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    (Original post by thegodofgod)
    (2y-1)(y+3) = 0

    y = \frac{1}{2} or y = 0
    Clearly not one of god's better days.
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    Please don't post full solutions :hmmm:
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    (Original post by ghostwalker)
    Clearly not one of god's better days.
    woops
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    (Original post by EierVonSatan)
    Please don't post full solutions :hmmm:
    But if I post it once, will it not help him (or her) the next time - so he (or she) doesn't have to post again for a similar question?

    On top of that, I'm new to using LaTeX, so I like to practise using it whenever I can :ninja: :getmecoat:
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    (Original post by thegodofgod)
    But if I post it once, will it not help him (or her) the next time - so he (or she) doesn't have to post again for a similar question?

    On top of that, I'm new to using LaTeX, so I like to practise using it whenever I can :ninja: :getmecoat:
    No, it encourages people to put up their homework and get it done for them and they learn little if anything. Guide them to the answer wherever possible :yes:
 
 
 
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