# solve a trigonometric equation

• Nov 10th 2009, 03:38 PM
satis
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.
• Nov 10th 2009, 04:16 PM
skeeter
Quote:

Originally Posted by satis
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 $\displaystyle \cos{x}$ ...

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

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

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

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

$\displaystyle x = \pi$
• Nov 10th 2009, 06:09 PM
pencil09
Quote:

Originally Posted by skeeter
$\displaystyle \sec{x} + \cos{x} = -2$

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 $\displaystyle 0<x<360$

Quote:

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 $\displaystyle 0<x<360$)
• Nov 10th 2009, 11:06 PM
mr fantastic
Quote:

Originally Posted by skeeter
$\displaystyle \sec{x} + \cos{x} = -2$

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.
• Nov 11th 2009, 05:52 AM
satis
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.
• Nov 12th 2009, 01:17 AM
pacman
yeah, experience plays a big role in it. Else, pure prowess . . . here is the graph of your equation.

http://www2.wolframalpha.com/Calcula...image/gif&s=10

(Bow)