# help...

• February 25th 2009, 02:17 PM
mathg33k
help...
ok so i got this problem and i did a bunch of things to try and solve it and i just cant seem to get it....

Cos(X)*Cos(2x)=1

your help would be appreciated. thanks.
• February 25th 2009, 04:04 PM
skeeter
Quote:

Originally Posted by mathg33k
ok so i got this problem and i did a bunch of things to try and solve it and i just cant seem to get it....

Cos(X)*Cos(2x)=1

your help would be appreciated. thanks.

well, it's not very easy to be done by hand ...

use the identity $\cos(2x) = 2\cos^2{x} - 1$

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

$2\cos^3{x} - \cos{x} - 1 = 0$

one obvious solution (and there is no other way to find it other than "observation") is $cos{x} = 1$

so, two solutions for x are $x = 0$ or $x = 2\pi$.

investigating other possible solutions ...

use synthetic division with the coefficients ...

Code:

1].........2..........0.........-1.........-1 .....................2...........2..........1 ----------------------------------------------- ...........2..........2..........1..........0
the depressed polynomial is ...

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

since $b^2-4ac < 0$, there are no other real solutions for $\cos{x}$

... told you it wasn't "nice".
• February 25th 2009, 04:18 PM
mathg33k
ok thanks that helps a lot.
• February 27th 2009, 11:19 PM
ekledes
notice

if cos (x)=1 this means x=0,360,2*360,.....
• February 28th 2009, 06:59 AM
skeeter
Quote:

Originally Posted by ekledes
notice

if cos (x)=1 this means x=0,360,2*360,.....

note that the given solutions were in the interval $0$ to $2\pi$ radians.

if you're trying to list possible solutions in degrees outside of that primary interval, then list them all ...

$x \in \{... \, , \, -2\cdot 360, -360, 0 , 360, 2 \cdot 360, \, ... \, \}$