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    x/2=x-x2 Im trying to find x

    Cant work it out, forgotten most of my basic algebra

    Any explanation would be most appreciated!
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    x/2=x-x^2

    x = 2(x-x^2)

    x = 2x - 2x^2

    0 = x - 2x^2

    x - 2x^2 = 0

    x(1 - 2x) = 0

    Solve for x.
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    (Original post by SherlockHolmes)
    x/2=x-x^2

    x = 2(x-x^2)

    x = 2x - 2x^2

    0 = x - 2x^2

    x - 2x^2 = 0

    x(1 - 2x) = 0

    Solve for x.
    so x=0 or x=1/2 ?
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    (Original post by breakeven)
    so x=0 or x=1/2 ?
    Yes

    (Original post by Buttercream:))
    anyone been to an interview at university of bradford for pharmacy, wondering what questions they ask really nervous interviews soon!
    Wrong thread I think!
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    Thanks everyone!
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    (Original post by SherlockHolmes)
    Yes



    Wrong thread I think!
    oh awkward -_- lmao! new to this my bad
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    any insight into this question?

    sinθtanθ=cosθ+1
    Show this can be expressed in the form: 2cos2θ+cosθ-1=0

    I took everything over to one side to get sinθtanθ -cosθ-1=0
    Then i thought I should use the trig identity to get
    sinθ(Sinθ/cosθ)-cosθ-1=0

    Kind of stuck from here, do the sinθ's cancel out or is it sin2θ?
    Also, what is sinθxcosθ ?
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    (Original post by breakeven)
    any insight into this question?

    sinθtanθ=cosθ+1
    Show this can be expressed in the form: 2cos2θ+cosθ-1=0

    I took everything over to one side to get sinθtanθ -cosθ-1=0
    Then i thought I should use the trig identity to get
    sinθ(Sinθ/cosθ)-cosθ-1=0

    Kind of stuck from here, do the sinθ's cancel out or is it sin2θ?
    Also, what is sinθxcosθ ?
    Multiply through by cosθ then change the sin2θ
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    (Original post by TenOfThem)
    Multiply through by cosθ then change the sin2θ
    How do I go about doing that?

    Im at sinθx(sinθ/cosθ)-cosθ-1=0
    Im trying to multiply out the brackets at the minute.
    I would get sin2θ+(sinθxcosθ)-cosθ-1=0

    Not sure what (sinθxcosθ) is at the minute but I would then change sin2θ into (1-cosθ)
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    (Original post by breakeven)
    How do I go about doing that?

    Im at sinθx(sinθ/cosθ)-cosθ-1=0
    Im trying to multiply out the brackets at the minute.
    I would get sin2θ+(sinθxcosθ)-cosθ-1=0

    Not sure what (sinθxcosθ) is at the minute but I would then change sin2θ into (1-cosθ)
    From here:
    sinθx(sinθ/cosθ)-cosθ-1=0

    When you multiply out the brackets, you should get sin2θ/cosθ.
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    (Original post by SherlockHolmes)
    From here:
    sinθx(sinθ/cosθ)-cosθ-1=0

    When you multiply out the brackets, you should get sin2θ/cosθ.
    So Im at sin2θ/cosθ-cosθ-1=0
    Now sure what to do from here, I need to make it the same as 2cos2θ+cosθ-1=0
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    (Original post by breakeven)
    So Im at sin2θ/cosθ-cosθ-1=0
    Now sure what to do from here, I need to make it the same as 2cos2θ+cosθ-1=0
    See TenOfThem's post above.
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    (Original post by SherlockHolmes)
    See TenOfThem's post above.
    I've tried that but I'm not getting the right answer? Sorry about this, must be driving you crazy but I do appreciate it!
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    (Original post by breakeven)
    I've tried that but I'm not getting the right answer? Sorry about this, must be driving you crazy but I do appreciate it!
    Post your workings so I can see where you are going wrong
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    (Original post by SherlockHolmes)
    Post your workings so I can see where you are going wrong
    Man I wish I could just take a picture of my notes haha.

    Okay so, sin2θ(sinθ/cosθ)-cosθ-1=0

    sin2θ x cosθ -cosθ-1=0
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    (Original post by breakeven)
    any insight into this question?

    sinθtanθ=cosθ+1
    Show this can be expressed in the form: 2cos2θ+cosθ-1=0

    I took everything over to one side to get sinθtanθ -cosθ-1=0
    Then i thought I should use the trig identity to get
    sinθ(Sinθ/cosθ)-cosθ-1=0

    Kind of stuck from here, do the sinθ's cancel out or is it sin2θ?
    Also, what is sinθxcosθ ?
    \sin \theta \tan \theta = \cos \theta + 1

    \sin \theta \dfrac{\sin \theta}{\cos \theta} = \cos \theta + 1

    \dfrac{\sin ^2 \theta}{\cos \theta} = \cos \theta + 1

    As I said Multiply by \cos \theta
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    (Original post by TenOfThem)
    \sin \theta \tan \theta = \cos \theta + 1

    \sin \theta \dfrac{\sin \theta}{\cos \theta} = \cos \theta + 1

    \dfrac{\sin ^2 \theta}{\cos \theta} = \cos \theta + 1

    As I said Multiply by \cos \theta
    Oh, god I get it now. Im so sorry!
    I wasn't multiplying the 1 by cosθ and i got confused, sorry and thanks so much!
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    (Original post by breakeven)
    Man I wish I could just take a picture of my notes haha.

    Okay so, sin2θ(sinθ/cosθ)-cosθ-1=0

    sin2θ x cosθ -cosθ-1=0
    sin^2θ/cosθ - when you multiply by cosθ you just get sin^2]θ.
    cosθ + 1 -when you multiply that by cosθ you get cos^2θ + cosθ.
    Now using the identity sin^2θ + cos^2θ = 1 you can change the sin^2θ into 1-cos^2θ (take this to the other side)
    0=cos^2-1+cos^2θ+cosθ (the original information from this side)
    Group like terms 2cos^2θ+cosθ-1
 
 
 
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