[SOLVED] Trig equation

**JannetheSwede** · July 4th 2009, 05:25 AM

Hello

Im having problems to solve the following trig equation.

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

Im given the following information

$x_{1}=2n\pi$

$x_{2}=\frac{a\pi }{b}+2n\pi$

$x_{3}=\frac{c\pi }{d}+2n\pi$

0 < $\frac{a\pi }{b}+2n\pi$ < $\frac{c\pi }{d}+2n\pi$ < $2\pi$

where n are integers

im not allowed to use a calculator, so im unsure how to eaven start...

**malaygoel** · July 4th 2009, 06:26 AM

What we have to solve or find here?

**JannetheSwede** · July 4th 2009, 06:33 AM

I have to find a, b, c and d

$x_{2}=\frac{a\pi }{b}+2n\pi$

$x_{3}=\frac{c\pi }{d}+2n\pi$

**skeeter** · July 4th 2009, 06:38 AM

Originally Posted by JannetheSwede

Im having problems to solve the following trig equation.

cos x-(1/2)/cos x=1/2

$\frac{\cos{x} - \frac{1}{2}}{\cos{x}} = \frac{1}{2}$

$\frac{2\cos{x} - 1}{\cos{x}} = 1$

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

$\cos{x} = 1$

$x = 2k\pi$ , $k \in \mathbb{Z}$

**JannetheSwede** · July 4th 2009, 06:39 AM

Originally Posted by skeeter

$\frac{\cos{x} - \frac{1}{2}}{\cos{x}} = \frac{1}{2}$

$\frac{2\cos{x} - 1}{\cos{x}} = 1$

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

$\cos{x} = 1$

$x = 2k\pi$ , $k \in \mathbb{Z}$

Yea thats one of three awnsers, but what i'm looking for are the other two... :S

btw, just to clearify $cosx-\frac{\frac{1}{2}}{cosx}=\frac{1}{2}$

**skeeter** · July 4th 2009, 06:43 AM

Originally Posted by JannetheSwede

Yea thats one of three awnsers, but what i'm looking for are the other two... :S

what makes you think there are more solutions?

**JannetheSwede** · July 4th 2009, 06:45 AM

ok, i have changed the initial question to clearify what it was

**skeeter** · July 4th 2009, 06:53 AM

Originally Posted by JannetheSwede

ok, i have changed the initial question to clearify what it was

$\cos{x} - \frac{\frac{1}{2}}{\cos{x}} = \frac{1}{2}<br />$

multiply every term by $2\cos{x}$ ...

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

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

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

$\cos{x} = -\frac{1}{2}$ ... $\cos{x} = 1$

you finish up.

**malaygoel** · July 4th 2009, 06:56 AM

$2cos^2x-cosx-1=0$

$cosx=1,\frac{-1}{2}$

Spoiler:

**JannetheSwede** · July 4th 2009, 07:06 AM

Ummh, or no... I still don't get it

I tried to apply this on another problem of the same type

$cosx+\frac{\frac{1}{2}}{cosx}=\frac{3}{2}$

but I was unable to solve it...

I did like this

$2cos^{2}x+cos x=3cos x$
$2cos^{2}x-2cos x=0$

and thats how far I got... what do I do from there?

**mathaddict** · July 4th 2009, 08:38 AM

Originally Posted by JannetheSwede

Ummh, or no... I still don't get it

I tried to apply this on another problem of the same type

$cosx+\frac{\frac{1}{2}}{cosx}=\frac{3}{2}$

but I was unable to solve it...

I did like this

$2cos^{2}x+cos x=3cos x$
$2cos^{2}x-2cos x=0$

and thats how far I got... what do I do from there?

you are wrong .

It simplifies to

$2cos^2x-3cosx+1=0$

$(2cosx-1)(cosx-1)=0$

$cosx=\frac{1}{2}$ , $cosx=1$

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