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    solve, for 0<x<2pi

    2sinXcos2X + sin2X = 0

    thanks
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    (Original post by ShOcKzZ)
    solve, for 0<x<2pi

    2sinXcos2X + sin2X = 0

    thanks
    cos2x=cos^2 (x) - sin^2 (x)

    2sinx(1 - 2sin^2 (x)) + sin2x = 0

    2sinx - 4sin^2 (x) + sin2x = 0

    3sinx - 4sin^2 (x) = 0

    3sinx = 4sin^2 (x)

    3 = 4sinx
    sinx = 3/4

    I think that's right.

    G
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    (Original post by gzftan)
    cos2x=cos^2 (x) - sin^2 (x)

    2sinx(1 - 2sin^2 (x)) + sin2x = 0

    2sinx - 4sin^2 (x) + sin2x = 0

    3sinx - 4sin^2 (x) = 0

    3sinx = 4sin^2 (x)

    3 = 4sinx
    sinx = 3/4

    I think that's right.

    G
    3rd line down, its '2sinx - 4sin^3 (x) + sin2x = 0
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    (Original post by gzftan)
    cos2x=cos^2 (x) - sin^2 (x)

    2sinx(1 - 2sin^2 (x)) + sin2x = 0

    2sinx - 4sin^2 (x) + sin2x = 0

    3sinx - 4sin^2 (x) = 0

    3sinx = 4sin^2 (x)

    3 = 4sinx
    sinx = 3/4

    I think that's right.

    G
    nup,

    it's not possible to do " 2sinX + sin2X = 3sinX"
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    (Original post by *dave*)
    3rd line down, its '2sinx - 4sin^3 (x) + sin2x = 0
    Sorry....got confused....it's easier on paper than on pc!!!!

    G
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    (Original post by ShOcKzZ)
    nup,

    it's not possible to do " 2sinX + sin2X = 3sinX"
    Is this P2 or higher? I know there isnt this type of trig in P3, but I havnt dun P4+ so I wouldnt be able to help u lol.

    Sorry
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    (Original post by *dave*)
    Is this P2 or higher? I know there isnt this type of trig in P3, but I havnt dun P4+ so I wouldnt be able to help u lol.

    Sorry
    I do double angle formulae in P3..maybe its different exam boards. I still can't figure it out yet though...and i have a p3 test 2moz!!!!

    G
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    (Original post by ShOcKzZ)
    solve, for 0<x<2pi

    2sinXcos2X + sin2X = 0

    thanks
    I think this is right:

    2sinxcos2x + sin2x = 0

    2sinx(2cos^2x -1) + 2sinxcosx = 0

    2sinx(2cos^2x -1 + cosx) = 0

    2sinx(2cosx +1)(cosx-1) = 0

    2sinx=0 which gives no solutions

    cosx = -1/2 and 1

    x = 120,360,0,240

    But because its between 0 and 360 the only valid answers are 120 and 240
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    Hmmmmm.....i've been thinking....how about this??

    2sinxcos2x + sin2x = 0

    2sinxcos2x + 2sinxcosx = 0

    [2sinx cancels out]

    cos2x + cosx = 0

    2cos^2 (x) - 1 + cosx = 0

    OR 2cos^2 (x) + cosx - 1 = 0

    Then factorise.

    G
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    Is this from Maths/Further Maths A-Level?
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    (Original post by gzftan)
    Hmmmmm.....i've been thinking....how about this??

    2sinxcos2x + sin2x = 0

    2sinxcos2x + 2sinxcosx = 0

    [2sinx cancels out]

    cos2x + cosx = 0

    cos^2 (x) - 1 + cosx = 0

    OR cos^2 (x) + cosx - 1 = 0

    Then use the formula to find two values for cosx....and work it out from there.

    G
    You cant just 'cancel out' you need to factorise, otherwise you can lose solutions..
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    (Original post by imasillynarb)
    I think this is right:

    2sinxcos2x + sin2x = 0

    2sinx(2cos^2x -1) + 2sinxcosx = 0

    2sinx(2cos^2x -1 + cosx) = 0

    2sinx(2cosx +1)(cosx-1) = 0

    2sinx=0 which gives no solutions

    cosx = -1/2 and 1

    x = 120,360,0,240

    But because its between 0 and 360 the only valid answers are 120 and 240
    Youve got the brackets the wrong way round ... its (2sinx)(2cosx - 1)(cosx + 1)=0

    2sinx=0 does give solutions lol ... what about zero and 2pi ?
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    (Original post by imasillynarb)
    You cant just 'cancel out' you need to factorise, otherwise you can lose solutions..
    Good point....i'd do well to remember that...cheers

    G
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    I think between us weve just about managed it lol
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    (Original post by *dave*)
    Youve got the brackets the wrong way round ... its (2sinx)(2cosx - 1)(cosx + 1)=0

    2sinx=0 does give solutions lol ... what about zero and 2pi ?
    Yeh, **** factorising by me, so now your solutions are, 0, 360, 300, 180, and 60...60, 180, 300 are the ones youd quote though...

    And yeh 2sinx=0 does have solutions, however I just put sin-1 0 into the calculator and got 0 so just thought 'bugger it doesnt have any'...im not thinking straight right now, allow me!
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    (Original post by *dave*)
    I think between us weve just about managed it lol
    Lol!! Now if only we could take the P3 exam this summer together!!!!

    We'd get it them all right!!! Or rather...you guys would...and i'd tag along!!!

    G
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    It works out at cosx = 1/2 and cosx = -1....... its easy from here!




    and the server is ****.
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    Yeh the server is a total disgrace..
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    Right, here goes...

    2sinXcos2X + sin2X = 0
    2SinX(Cos^2X - Sin^2X) + 2SinxCosX=0
    2SinX(Cos^2X - Sin^2X + CosX) = 0
    2SinX(Cos^2X - (1- Cos^2X) + CosX) = 0
    2SinX(2Cos^2X + CosX -1) = 0
    2SinX(2CosX + 1)(CosX - 1) = 0

    SinX = 0 --> X = 0, 180, 360
    CosX = 1 --> X = 0 or 360.
    CosX = -1/2 ---> X = 120, 240.

    answers: 0, 120, 180, 240, 360.
    or in radians: 0, 2pi/3, pi, 4pi/3, 2pi.

    did that work out right? Someone check my algebra...i make silly mistakes often.
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    Ralfskini is wrong.

    dont be startled its an easy error-

    it works out at 2cosx= 1/2

    and therefore sin^2-cos= 1/2

    factorise from here and bob's your uncles brother.
 
 
 
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