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.

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- Feb 25th 2009, 02:17 PMmathg33khelp...
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. - Feb 25th 2009, 04:04 PMskeeter
well, it's not very easy to be done by hand ...

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

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

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

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

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

investigating other possible solutions ...

use synthetic division with the coefficients ...

Code:`1].........2..........0.........-1.........-1`

.....................2...........2..........1

-----------------------------------------------

...........2..........2..........1..........0

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

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

... told you it wasn't "nice". - Feb 25th 2009, 04:18 PMmathg33k
ok thanks that helps a lot.

- Feb 27th 2009, 11:19 PMekledes
notice

if cos (x)=1 this means x=0,360,2*360,..... - Feb 28th 2009, 06:59 AMskeeter
note that the given solutions were in the interval $\displaystyle 0$ to $\displaystyle 2\pi$ radians.

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

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