Posted by Kate on Tuesday, July 27, 2010 at 2:34pm.
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
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(4^2-4(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 +/-
ln(-2 + sqrt(3)) /i = x
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
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