solve a trigonometric equation

**satis** · November 10th 2009, 03:38 PM

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.

**skeeter** · November 10th 2009, 04:16 PM

Originally Posted by satis

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 $\cos{x}$ ...

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

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

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

$\cos{x} = -1$

$x = \pi$

**pencil09** · November 10th 2009, 06:09 PM

Originally Posted by skeeter

$\sec{x} + \cos{x} = -2$

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 $0<x<360$ )

**mr fantastic** · November 10th 2009, 11:06 PM

Originally Posted by skeeter

$\sec{x} + \cos{x} = -2$

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.

**satis** · November 11th 2009, 05:52 AM

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.

**pacman** · November 12th 2009, 01:17 AM

yeah, experience plays a big role in it. Else, pure prowess . . . here is the graph of your equation.

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