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Math Help - a few questions from homework

  1. #1
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    a few questions from homework

    I have to find all answers to these on the interval 0-2 pi, I think I'm getting them right but I'm not 100% positive

    For the first one I had
    cos x-2sinxcosx=0

    and I did
    (2cosx-2sinxcosx)/cosx=0
    1-2sinx=0
    1=2sinx
    sinx=1/2

    and on the interval 0-2pi I got
    pi/6, 5pi/6

    the second one I have to do
    2sin^2 x=4sinx+6

    and I got
    sin^2 x=2sinx+3
    sin^2 x-2sinx-3=0
    x^2-2x-3=0
    (x-3)(x+1)
    sinx=3 sinx=-1

    and for the interval I got
    3pi/2

    are there any answers on the interval that I am missing?

    also if anyone can help me with this identity
    (cscx-cotx)^2=(1-cosx)/(1+cosx)

    I've tried everything but have gotten nowhere
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  2. #2
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    Quote Originally Posted by prerunner View Post
    I have to find all answers to these on the interval 0-2 pi, I think I'm getting them right but I'm not 100% positive

    For the first one I had
    cos x-2sinxcosx=0

    and I did
    (2cosx-2sinxcosx)/cosx=0 Mr F says: Wrong. You have to factorise:

    {\color{red}\cos x (1 - 2 \sin x) = 0 \Rightarrow \cos x = 0} or {\color{red}1 - 2 \sin x = 0}.

    1-2sinx=0
    1=2sinx
    sinx=1/2

    and on the interval 0-2pi I got
    pi/6, 5pi/6 Mr F says: For reasons given above, you're missing solutions.

    the second one I have to do
    2sin^2 x=4sinx+6

    and I got
    sin^2 x=2sinx+3
    sin^2 x-2sinx-3=0
    x^2-2x-3=0 Mr F advises: Use a different symbol to x. x is already being used in the question. Use w = sin x so that w^2 - 2w - 3 = 0.

    (x-3)(x+1)
    sinx=3 sinx=-1

    and for the interval I got
    3pi/2

    are there any answers on the interval that I am missing? Mr F says: No.

    [snip]
    ..
    Last edited by mr fantastic; November 17th 2008 at 10:03 PM.
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  3. #3
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    Quote Originally Posted by prerunner View Post
    [snip]
    also if anyone can help me with this identity
    (cscx-cotx)^2=(1-cosx)/(1+cosx)

    I've tried everything but have gotten nowhere
    Left Hand Side = (\text{cosec} \, x - \text{cot}\, x)^2

    = \left(\frac{1}{\sin x} - \frac{\cos x}{\sin x}\right)^2

    = \left(\frac{1 - \cos x}{\sin x}\right)^2

    = \frac{(1 - \cos x)^2}{\sin^2 x}

    = \frac{(1 - \cos x)^2}{1 - \cos^2 x}

    = \frac{(1 - \cos x)^2}{(1 - \cos x)(1 + \cos x)}

    = ....
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