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Math Help - solving equations help

  1. #1
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    solving equations help

    given that -\pi <= x <= \pi

    solve:

    a) cos^2x+cosx=0

    b) 2sin^2x - sinx = 0


    NOTE: <= (less than or equal to)


    any help much appreciated!
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  2. #2
    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by jvignacio View Post
    given that -\pi <= x <= \pi

    solve:

    a) cos^2x+cosx=0

    b) 2sin^2x - sinx = 0


    NOTE: <= (less than or equal to)


    any help much appreciated!
    cos(x)(1+cos(x)) = 0

    Either cos(x) = 0 or 1+cos(x) = 0

    <br />
cos(x) = 0~ when~ x = \pi/2 ~or~ - \pi/2

    cos(x) = -1~ when ~x= \pi~ or~ x = - \pi

    2sin^2(x) - sin(x) = 0

    sin(x) (2sin(x)-1) = 0

    either sin(x) = 0 or sin(x) = 1/2

    sin(x) = 0 ~when ~ x = 0 ~or ~x= \pi ~or~x= -\pi
    <br />
sin(x) = 1/2 ~when ~x = \pi/6~ or ~x = 5\pi/6
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  3. #3
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    Quote Originally Posted by jvignacio View Post
    given that -\pi <= x <= \pi

    solve:

    a) cos^2x+cosx=0

    b) 2sin^2x - sinx = 0


    NOTE: <= (less than or equal to)


    any help much appreciated!
    You can factorise both of these:

    cos(x)(cos(x) + 1) = 0 so cos(x) = 0 or cos(x) = -1

    and the same with sin(x). Just remember the points at -1 will repeat every +-[tex]2\pi/MATH] and the points at 0 will repeat every +- \pi
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  4. #4
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    Quote Originally Posted by ADARSH View Post
    cos(x)(1+cos(x)) = 0

    Either cos(x) = 0 or 1+cos(x) = 0

    <br />
cos(x) = 0~ when~ x = \pi/2 ~or~ - \pi/2

    cos(x) = -1~ when ~x= \pi~ or~ x = - \pi

    2sin^2(x) - sin(x) = 0

    sin(x) (2sin(x)-1) = 0

    either sin(x) = 0 or sin(x) = 1/2

    sin(x) = 0 ~when ~ x = 0 ~or ~x= \pi ~or~x= -\pi
    <br />
sin(x) = 1/2 ~when ~x = \pi/6~ or ~x = 5\pi/6
    Hi, how did you simplify cos^2x+cosx ?
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  5. #5
    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by jvignacio View Post
    Hi, how did you simplify cos^2x+cosx ?

    cos^2(x) + cos(x) = cos(x)\times cos(x) + cos(x)

    Take cos(x) as common thus

    cos(x)*(cos(x)+1) = 0

    Now either of the two terms is zero

    so cos(x) = 0 or cos(x) = -1
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  6. #6
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    Quote Originally Posted by ADARSH View Post

    cos^2(x) + cos(x) = cos(x)\times cos(x) + cos(x)

    Take cos(x) as common thus

    cos(x)*(cos(x)+1) = 0

    Now either of the two terms is zero

    so cos(x) = 0 or cos(x) = -1
    okay so you have cos(x) = 0 and cos(x) = -1

    now what do u do with the -\pi <= x <= \pi ?
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  7. #7
    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by jvignacio View Post
    okay so you have cos(x) = 0 and cos(x) = -1

    now what do u do with the -\pi <= x <= \pi ?
    When is cos(x) = 0 or when is cos(x) = -1

    What are the values of x when this values are satisfied

    For x= 0 This happens when x= 90degree or x = - 90 degrees
    And for cos(x)= -1 this happens when x= 180degree or x= -180 degree

    This is done using the property that
    cos(-x) =cos(x)
    ie; if the answer is 90 degree than answer is also -90 degree as this value also lies in the domain

    Read this
    http://en.wikipedia.org/wiki/Trigonometric_function
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