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Math Help - Consider the equation 2cos^2x+sinx-1=0

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    Consider the equation 2cos^2x+sinx-1=0

    Consider the equation 2cos^2x + sinx - 1 = 0

    a) Explain why the equation cannot be factored.

    b) Suggest a trigonometric identity that can be used to remove the problem identified in part a).

    c) Apply the identity and rearrange the equation into a factorable form.

    d) Factor the equation.

    e) Determine all solutions in the interval x E [0,2pi].
    Last edited by kmjt; December 18th 2009 at 03:53 PM.
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  2. #2
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    Quote Originally Posted by kmjt View Post
    Ok.. so I tried to factor it. This is my steps:

    2cos^2x + sinx - 1 =0

    Err i change the way it looks to make it easier for me

    2x^2 + x - 1 =0

    2x^2 + 2x -1x -1

    (2x^2 + 2x) + (-1x - 1)

    2x(x+1) -1(x+1)

    (2x-1)(x+1)

    (2cosx - 1)(cosx + 1)

    Isn't that factoring it? Why is it asking why you cannot factor it? What am I doing wrong?
    hmm... A silly mistake

    The equation is 2cos^2x + sinx - 1 = 0 and not 2cos^2x + cosx - 1 =0
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    Quote Originally Posted by kmjt View Post
    Ok.. so I tried to factor it. This is my steps:

    2cos^2x + sinx - 1 =0

    Err i change the way it looks to make it easier for me

    2x^2 + x - 1 =0

    ...

    Isn't that factoring it? Why is it asking why you cannot factor it? What am I doing wrong?
    It looks like you've substituted x for both cos x and sin x. That's not good.

    The way to proceed is to use a suitable trigonometric identity ... say one involving squares of cosines and squares of sines, somehow ... to transform your cos squared term into a sin squared term.
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    Quote Originally Posted by kmjt View Post
    Consider the equation 2cos^2x + sinx - 1 = 0

    a) Explain why the equation cannot be factored.

    b) Suggest a trigonometric identity that can be used to remove the problem identified in part a).

    c) Apply the identity and rearrange the equation into a factorable form.

    d) Factor the equation.

    e) Determine all solutions in the interval x E [0,2pi].

    Ok.. so I tried to factor it. This is my steps:

    2cos^2x + sinx - 1 =0

    Err i change the way it looks to make it easier for me

    2x^2 + x - 1 =0

    2x^2 + 2x -1x -1

    (2x^2 + 2x) + (-1x - 1)

    2x(x+1) -1(x+1)

    (2x-1)(x+1)

    (2cosx - 1)(cosx + 1)

    Isn't that factoring it? Why is it asking why you cannot factor it? What am I doing wrong?
    you cannot represent \cos{x} and \sin{x} with the same variable.


    2\cos^2{x} + \sin{x} - 1 = 0<br />

    change cos^2{x} to 1 - \sin^2{x} and try again.
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    I see! I'm blind I didn't notice that it was sin :$ Thanks guys.
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    Ok so i'm on part e.. actually factoring it. This is what I come up with:

    2cos^2x+sinx-1=0

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

    2-2sin^2x+sinx-1=0

    -2sin^2x+sinx-1+2=0

    -2sin^2x+sinx+1=0

    So I try to factor it..

    (-2x^2+2x)+(-1x+1)

    so that looks pretty wrong..

    This is what I found on yahoo answers when someone was doing a similar problem:

    You know cos²x = 1 - sin²x, so:
    2cos²x - sinx - 1 = 0 → 2 (1 - sin²x) - sinx - 1 = 0 →
    2 - 2sin²x - sinx - 1 = 0 →
    - 2sin²x - sinx + 1 = 0 → 2sin²x + sinx - 1 = 0

    How does he go from -2sin^2x to it being positive 2sin^2x, and how come the -sinx all of a sudden switches to +sinx? And the +1 switches to -1? Did he just bring everything on to the other side of the equal sign?
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    Quote Originally Posted by kmjt View Post
    Ok so i'm on part e.. actually factoring it. This is what I come up with:

    2cos^2x+sinx-1=0

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

    2-2sin^2x+sinx-1=0

    -2sin^2x+sinx-1+2=0

    -2sin^2x+sinx+1=0

    So I try to factor it..

    (-2x^2+2x)+(-1x+1)

    so that looks pretty wrong..

    This is what I found on yahoo answers when someone was doing a similar problem:

    You know cos²x = 1 - sin²x, so:
    2cos²x - sinx - 1 = 0 → 2 (1 - sin²x) - sinx - 1 = 0 →
    2 - 2sin²x - sinx - 1 = 0 →
    - 2sin²x - sinx + 1 = 0 → 2sin²x + sinx - 1 = 0

    How does he go from -2sin^2x to it being positive 2sin^2x, and how come the -sinx all of a sudden switches to +sinx? And the +1 switches to -1? Did he just bring everything on to the other side of the equal sign?
    yes ... or you can look at it as multiplying both sides by (-1)
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    Can anyone clarify if my answers are right?

    c) 2cos^2x+sinx-1=0

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

    2-2sin^2x+sinx-1=0

    -2sin^2x+sinx-1+2=0

    -2sin^2x+sinx+1=0

    2sin^2x-sinx-1=0


    d) 2sin^2x-sinx-1=0

    and from that I got to

    (2sinx+1)(sinx-1)

    e) The solutions in the interval [0,2pi] are pi/2, 7pi/6 and 11pi/6.

    Is that right?
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    Quote Originally Posted by kmjt View Post
    Can anyone clarify if my answers are right?

    c) 2cos^2x+sinx-1=0

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

    2-2sin^2x+sinx-1=0

    -2sin^2x+sinx-1+2=0

    -2sin^2x+sinx+1=0

    2sin^2x-sinx-1=0


    d) 2sin^2x-sinx-1=0

    and from that I got to

    (2sinx+1)(sinx-1)

    e) The solutions in the interval [0,2pi] are pi/2, 7pi/6 and 11pi/6.

    Is that right?
    Yes.
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