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Thread: 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

    $\displaystyle secx + cosx = -2$

    using identities, I can turn that into

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

    and from there

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

    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

    $\displaystyle secx + cosx = -2$

    using identities, I can turn that into

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

    and from there

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

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

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

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

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

    $\displaystyle (\cos{x}+1)^2 = 0
    $

    $\displaystyle \cos{x} = -1$

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

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

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

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

    $\displaystyle (\cos{x}+1)^2 = 0
    $

    $\displaystyle \cos{x} = -1$

    $\displaystyle x = \pi$

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

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

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

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

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

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

    $\displaystyle (\cos{x}+1)^2 = 0
    $

    $\displaystyle \cos{x} = -1$

    $\displaystyle x = \pi$
    Since no domain is specified, I'd be inclined to give the solution as either $\displaystyle x = \pi + 2n \pi = (2n + 1) \pi$ (working in radians) or $\displaystyle x = 180 + 360 n = 180 (2n + 1)$ (working in degrees) where $\displaystyle n$ 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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