help...

mathg33k · #1 $Old$ February 25th, 2009, 03:17 PM

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.

skeeter · #2 $Old$ February 25th, 2009, 05:04 PM

Quote:

Originally Posted by mathg33k $View Post$

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".

mathg33k · #3 $Old$ February 25th, 2009, 05:18 PM

ok thanks that helps a lot.

ekledes · #4 $Old$ February 28th, 2009, 12:19 AM

notice

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

skeeter · #5 $Old$ February 28th, 2009, 07:59 AM

Quote:

Originally Posted by ekledes $View Post$

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 ...

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