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December 19, 2014

December 19, 2014

Posted by **Kate** on Tuesday, July 27, 2010 at 6:44pm.

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

- CALC -
**bobpursley**, Tuesday, July 27, 2010 at 7:10pmok, 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

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