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Thread: 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
    $\displaystyle cos x-2sinxcosx=0$

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

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

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

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

    and for the interval I got
    $\displaystyle 3pi/2$

    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
    $\displaystyle cos x-2sinxcosx=0$

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

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

    $\displaystyle 1-2sinx=0$
    $\displaystyle 1=2sinx$
    $\displaystyle sinx=1/2$

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

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

    and I got
    $\displaystyle sin^2 x=2sinx+3$
    $\displaystyle sin^2 x-2sinx-3=0$
    $\displaystyle x^2-2x-3=0$ 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.

    $\displaystyle (x-3)(x+1)$
    $\displaystyle sinx=3 sinx=-1$

    and for the interval I got
    $\displaystyle 3pi/2$

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

    [snip]
    ..
    Last edited by mr fantastic; Nov 17th 2008 at 09:03 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 $\displaystyle = (\text{cosec} \, x - \text{cot}\, x)^2$

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

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

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

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

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

    = ....
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