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Math Help - solve a trigonometric equation

  1. #1
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    solve a trigonometric equation

    could I get some assistance solving this equation? I'm afraid I'm just not getting anywhere with it

    secx + cosx = -2

    using identities, I can turn that into

    \frac{1}{cosx} + cosx = -2

    and from there

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

    unfortunately I peter out there, assuming I didn't make some mistake in that process. Any pointers would be appreciated.
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  2. #2
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    Quote Originally Posted by satis View Post
    could I get some assistance solving this equation? I'm afraid I'm just not getting anywhere with it

    secx + cosx = -2

    using identities, I can turn that into

    \frac{1}{cosx} + cosx = -2

    and from there

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

    unfortunately I peter out there, assuming I didn't make some mistake in that process. Any pointers would be appreciated.
    \sec{x} + \cos{x} = -2

    multiply every term by \cos{x} ...

    1 + \cos^2{x} = -2\cos{x}

    \cos^2{x} + 2\cos{x} + 1 = 0

    (\cos{x}+1)^2 = 0<br />

    \cos{x} = -1

    x = \pi
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  3. #3
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    Quote Originally Posted by skeeter View Post
    \sec{x} + \cos{x} = -2

    multiply every term by \cos{x} ...

    1 + \cos^2{x} = -2\cos{x}

    \cos^2{x} + 2\cos{x} + 1 = 0

    (\cos{x}+1)^2 = 0<br />

    \cos{x} = -1

    x = \pi

    i think is depend on the interval of x...
    if 0<x<360

    and use this formula
    if \cos {x}= \cos {a}
    x=a + k (360) or x=-a+k(360)
    in this case:
    \cos{x} = -1
    so, \cos{x} = cos{180}

    x=180 + k (360)
    let  x=0 so x=180 + (0) (360)= 180
    let  x=1 so x=180 + (1) (360)= 540 (rejected, bcause 540 not in the interval 0<x<360)
    or
    x=-a+k(360)
    let  x=0 so x=-180 + (0) (360)= -180 (rejected, bcause -180 not in the interval 0<x<360)
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  4. #4
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    Quote Originally Posted by skeeter View Post
    \sec{x} + \cos{x} = -2

    multiply every term by \cos{x} ...

    1 + \cos^2{x} = -2\cos{x}

    \cos^2{x} + 2\cos{x} + 1 = 0

    (\cos{x}+1)^2 = 0<br />

    \cos{x} = -1

    x = \pi
    Since no domain is specified, I'd be inclined to give the solution as either x = \pi + 2n \pi = (2n + 1) \pi (working in radians) or x = 180 + 360 n = 180 (2n + 1) (working in degrees) where n is an integer.
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  5. #5
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    Thank you all for the insight. It definitely makes sense in retrospect, but hindsight is always 20/20.

    Is there any trick to seeing the correct path to go to solve this kind of stuff, or is it just a matter of experience? I'm afraid I may have a few more of this type of question in the near future.
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  6. #6
    Senior Member pacman's Avatar
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    yeah, experience plays a big role in it. Else, pure prowess . . . here is the graph of your equation.



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