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    solve :

    2sin2x=tanx

    for 0-360 deg
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    (Original post by ShOcKzZ)
    solve :

    2sin2x=tanx

    for 0-360 deg
    Use the double angle formula first to simplify left hand side....

    can't quite recall them...but its something like...

    sin2x = 2sinxcosx????

    G
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    (Original post by gzftan)
    Use the double angle formula first to simplify left hand side....

    can't quite recall them...but its something like...

    sin2x = 2sinxcosx????

    G
    Yeah, that's right.
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    (Original post by Nylex)
    Yeah, that's right.
    yes, so:

    2 [ 2sinxcosx ] = tanx

    4sinxcosx = sinx/cosx

    4sinxcos^2x = sinx

    4sinx(1-sin^2x) = sinx

    now??..
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    (Original post by ShOcKzZ)
    solve :

    2sin2x=tanx

    for 0-360 deg
    Is this P3 Maths Shockzz, yes?
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    (Original post by ShOcKzZ)
    yes, so:

    2 [ 2sinxcosx ] = tanx

    4sinxcosx = sinx/cosx

    4sinxcos^2x = sinx

    4sinx(1-sin^2x) = sinx

    now??..
    4sinx(1-sin^2x) - sinx = 0
    4sinx(1-sin^2x - 1/4) = 0.

    So sin(x) = 0

    or sin^2x = 3/4 so sin(x) = +/- root3/2.
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    (Original post by bono)
    Is this P3 Maths Shockzz, yes?
    p4 aqa B
    i think thats p3 work for most other boards.
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    2sin(2x) = tan(x)

    So, using the double angle formula:

    sin(2x) = sin(x+x) = sin(x)cos(x)+sin(x)cos(x) = 2sin(x)cos(x)

    => 2(2sin(x)cos(x)) = tan(x)

    We know that tan(x) = sin(x)/cos(x):

    4sin(x)cos(x) = sin(x)/cos(x)

    Cancel the sin(x):

    4cos(x) = 1/cos(x)
    4cos^2(x) = 1
    cos^2(x) = 1/4
    cos(x) = sqrt(1/4) = 1/2

    So,

    x = cos^-1(1/2) = 60 deg.
    OR x = 360 - 60 = 300 deg.

    I checked this numerically too.
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    (Original post by theone)
    4sinx(1-sin^2x) - sinx = 0
    4sinx(1-sin^2x - 1/4) = 0.

    So sin(x) = 0

    or sin^2x = 3/4 so sin(x) = +/- root3/2.
    amazing, i have repped you, thanks.
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    (Original post by capslock)
    2sin(2x) = tan(x)

    So, using the double angle formula:

    sin(2x) = sin(x+x) = sin(x)cos(x)+sin(x)cos(x) = 2sin(x)cos(x)

    => 2(2sin(x)cos(x)) = tan(x)

    We know that tan(x) = sin(x)/cos(x):

    4sin(x)cos(x) = sin(x)/cos(x)

    Cancel the sin(x):

    4cos(x) = 1/cos(x)
    4cos^2(x) = 1
    cos^2(x) = 1/4
    cos(x) = sqrt(1/4) = 1/2

    So,

    x = cos^-1(1/2) = 60 deg.
    OR x = 360 - 60 = 300 deg.

    I checked this numerically too.
    a) You can't cancel the sin(x) (what if they are 0?)

    b) You've forgotten to take the negative root of 1/4 into account as well.
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    (Original post by ShOcKzZ)
    amazing, i have repped you, thanks.
    Yes Shockzz I do the same syllabus!!

    P1, P2, M1, P4, P5, P2.
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    (Original post by ShOcKzZ)
    amazing, i have repped you, thanks.
    No prob, thanks
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    (Original post by bono)
    Yes Shockzz I do the same syllabus!!

    P1, P2, M1, P4, P5, P2.
    u do p2 twice?? hehe
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    (Original post by ShOcKzZ)
    p4 aqa B
    i think thats p3 work for most other boards.
    Yes, AQA's modules are a bit different.
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    (Original post by bono)
    Yes, AQA's modules are a bit different.
    yes, i do

    p1,p2,s1,p4,p5,m3
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    (Original post by theone)
    4sinx(1-sin^2x) - sinx = 0
    4sinx(1-sin^2x - 1/4) = 0.
    - sinx = -1/4

    :confused:
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    (Original post by ShOcKzZ)
    yes, i do

    p1,p2,s1,p4,p5,m3
    Why you gonna do M3 when you can do M2?

    And don't you need M1 to do a later Mechanics module?? :confused:
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    (Original post by bono)
    - sinx = -1/4

    :confused:
    Err, what? (1 - sin^2(x) - 1/4) = 0 <=> 3/4 - sin^2(x) = 0 <=> 3/4 = sin^2(x).
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    (Original post by bono)
    Why you gonna do M3 when you can do M2?

    And don't you need M1 to do a later Mechanics module?? :confused:
    I have no idea.
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    How the hell could AQA possibly justify leaving double angle formula off P3?
 
 
 
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