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(4^2-4(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 +/-
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
ok, where you did the two branches..
ln(-2+-sqrt3)/i=x
1)
ln(-2+sqrt3/i= -iln(-2+sqrt3)=x
2)
ln:(-2-sqrt3)/i=-iln(-2-sqrt3)=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
The mistake in your solution lies in the step where you set cos(x) equal to -2 and started solving further. The equation cos(x) = -2 does not have any real solutions. The range of the cosine function is [-1, 1], so it is not possible for cos(x) to be -2.
In your solution, you used complex numbers and the exponential form of trigonometric functions to arrive at a quadratic equation for cis(x). However, when you solved the quadratic equation, you obtained values for cis(x) which do not correspond to valid values of cos(x).
To solve cos(x) = -2, you can use the inverse cosine function. However, it is important to note that the inverse cosine function only gives values within the range [-π/2, π/2]. Hence, the equation cos(x) = -2 has no real solutions.