The solution to cos x = 2 or any number > 1

[smutangama]

Homework Statement

Generally the inverse cosine of any number > 1

Homework Equations

cos x = 2

The Attempt at a Solution

Obviously by putting this in a calculator, you get an error so the root has to be complex

I used the identity (e^iθ+e^-iθ)/2 = cosθ

through a bit of manipulation, I came to e^iθ = 2±√3

with the solutions being x = ln(2±√3) / i

is this correct?

thanks

 

[Electric Red]

Watch out with your calculations, to keep them clear for yourself and others.
But, your values for x are indeed correct.

 

[HallsofIvy]

Homework Statement

Generally the inverse cosine of any number > 1

Homework Equations

cos x = 2

The Attempt at a Solution

Obviously by putting this in a calculator, you get an error so the root has to be complex

I used the identity (e^iθ+e^-iθ)/2 = cosθ

through a bit of manipulation, I came to e^iθ = 2±√3

with the solutions being x = ln(2±√3) / i

is this correct?

thanks

Correct so far but you have not finished. You should be able to get the answer in eityer of the
"standard forms" a+ bi or [tex]re^{i\theta}[/tex]. First, dividing by i is the same as multiplying by -i so this is [itex]x= -ln(2\pm \sqrt{3})i[/itex]. Next, ln(x) where x is a NEGATIVE real number is itself complex.

 

[haruspex]

ln(x) where x is a NEGATIVE real number is itself complex.

True, but 2 - √3 > 0.