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

1. Aug 25, 2013

### smutangama

1. The problem statement, all variables and given/known data

Generally the inverse cosine of any number > 1

2. Relevant equations

cos x = 2

3. 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+e-iθ)/2 = cosθ

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

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

is this correct?

thanks

2. Aug 25, 2013

### Electric Red

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

3. Aug 25, 2013

### HallsofIvy

Staff Emeritus
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 $$re^{i\theta}$$. First, dividing by i is the same as multiplying by -i so this is $x= -ln(2\pm \sqrt{3})i$. Next, ln(x) where x is a NEGATIVE real number is itself complex.

4. Aug 25, 2013

### haruspex

True, but 2 - √3 > 0.