Posted by Kate on .
1. The problem statement, all variables and given/known data
cos(x) = 2
I was trying to solve that problem ignoring what people say about it not being defined and got a number strangely. I showed step by step so you can tell me what I did wrong as I don't think I did anything wrong what so ever.
2. Relevant equations
In work below
3. The attempt at a solution
e^(ix) = cosx + i sinx = cisx
e^(ix) = cosx  i sinx = cis(x)
cis(x) + cis(x) = cosx + i sin x + cosx  i sinx = 2cosx
from which
cosx = (cis(x) + cis(x) )/2
from which I set it equal to 2 and began to solve
cosx = (cis(x) + cis(x) )/2 = 2
Multipled by 2 on both sides
cis(x) + cis(x) ) = 4
Multipled by cisx on both sides
cis^2(x) + 1 = 4cis(x)
set equal to zero by adding 4cis(x) to both sides
cis^2(x) + 4cis(x) + 1 = 0
solved the quadratic for cis(x)
(4 +/ sqrt(4^24(1)))/2
= 2 +/ sqrt(16  4)/2
= 2 +/ sqrt(12)/2
= 2 +/ (2sqrt(3))/2
= 2 +/ sqrt(3)
cis(x) = 2 +/ sqrt(3) = e^(ix)
solved for x took natural log of both sides
ln(2 +/ sqrt(3)) = ln( e^(ix) )
took out exponents and used ln(e) = 1
ln(2 +/ sqrt(3)) = ix
divided through by i
ln(2 +/ sqrt(3)) /i = x
simpified for +/
for +
ln(2 + sqrt(3)) /i = x
for 
ln(2  sqrt(3)) /i
factored out negative one
ln((2 + sqrt(3))/i
used the fact that ln(xy) = lnx + lny
( ln(1) + ln(2 + sqrt(3)) )/i
used the fact that ln(1) = i pi
( i pi + ln(2 + sqrt(3)) )/i
canceled out the i
pi + ln((2 + sqrt(3)) )/i = x

CALC 
bobpursley,
ok, where you did the two branches..
ln(2+sqrt3)/i=x
1)
ln(2+sqrt3/i= iln(2+sqrt3)=x
2)
ln:(2sqrt3)/i=iln(2sqrt3)=x
Now consider that ln(z)=ln(1/z)
you can then rationalize the denominator to get the correct answer.
See http://www.physicsforums.com/showthread.php?p=2816568