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Math Help - Solutions with intervals

  1. #1
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    Solutions with intervals

    Find all solutions in the interval [0, 2pi)

    2sin^2 x = 2 + cox x

    I kno that most of the solutions are gonna be 1st quadrant solutions,

    sin is positive in the 1st and 2nd and cos is negative in the 4th quadrant
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  2. #2
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    Make these changes.
    \begin{array}{l} 2\sin ^2 (x) = 2 + \cos (x) \\  2 - 2\cos ^2 (x) = 2 + \cos (x) \\  2\cos ^2 (x) + \cos (x) = 0 \\  \end{array}

    Can you solve 2y^2  + y = 0?
    Let y = \cos (x).
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  3. #3
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    yes i could solve that thank you fo rthe asistance, i dont get where your getting
    2-2cos^2 x + 2 + cos(x)
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    so are these the solutions

    y=cos x

    (2y-1)(y+1)

    Cos(x)=1/2 and cos(x)=-1

    is this correct
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by NoAsherelol View Post
    yes i could solve that thank you fo rthe asistance, i dont get where your getting
    2-2cos^2 x + 2 + cos(x)
    he used the identity (that you should be familiar wit) \sin^2 x + \cos^2 x = 1. that means you can change \sin^2 x to 1 - \cos^2 x, which is what Plato did
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  6. #6
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    Quote Originally Posted by Jhevon View Post
    he used the identity (that you should be familiar wit) \sin^2 x + \cos^2 x = 1. that means you can change \sin^2 x to 1 - \cos^2 x, which is what Plato did
    is my factoring correct
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by NoAsherelol View Post
    is my factoring correct
    no

    2y^2 + y = 0

    \Rightarrow y(2y + 1) = 0


    by the way, you could check whether or not you factored right by multiplying out what you got. you would notice you end up with three terms, as opposed to two, which is what you started with
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    Quote Originally Posted by Jhevon View Post
    no

    2y^2 + y = 0

    \Rightarrow y(2y + 1) = 0


    by the way, you could check whether or not you factored right by multiplying out what you got. you would notice you end up with three terms, as opposed to two, which is what you started with

    Im coming up with cos(x)=0 and cos(x)=-1, but the interval is [0, 2pi) with those answers it could only be 1 solution but it asked me to find solutions yo i suck at this class, i might have tell my guidance counselor to switch me out.....sorry if im buiggin the mess out of you
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by NoAsherelol View Post
    Im coming up with cos(x)=0 and cos(x)=-1, but the interval is [0, 2pi) with those answers it could only be 1 solution but it asked me to find solutions yo i suck at this class, i might have tell my guidance counselor to switch me out.....sorry if im buiggin the mess out of you
    the solutions are cos(x) = 0 and cos(x) = -1/2... now, how would we find x?
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  10. #10
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    Quote Originally Posted by Jhevon View Post
    the solutions are cos(x) = 0 and cos(x) = -1/2... now, how would we find x?
    well i know cos is negative in the 2nd and fourth quadrants,

    so possible solutions are x=2pi/3 and x=11pi/6
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  11. #11
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by NoAsherelol View Post
    well i know cos is negative in the 2nd and fourth quadrants,

    so possible solutions are x=2pi/3 and x=11pi/6
    from the graph of cos(x) we know that \cos x = 0 \implies x = \frac {\pi}2 + k \pi for k \in \mathbb{Z}

    from the table i gave you in one of your other posts and your knowledge of reference angles, we know \cos x = - \frac 12 \implies x = \frac {2 \pi}3 + k \pi \mbox{ or } \frac {4 \pi}3 + n \pi for n,k \in \mathbb{Z}

    (cosine is negative in the second and third quadrants)
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