Question

How do you solve `cos x = 3` ?

Answer 1

`"no solution"`

Explanation:

`"note that "-1<=cosx <=1`

`rArrcosx=3" is not valid"`

Answer 2

`x` has no value.

Explanation:

The cosine function only gives out answers from `-1` to `1`. As this value exceeds these bounds, `x` has no value.

Answer 3

`x = 2npi+-iln(3+2sqrt(2))" "` for any `n in ZZ`

Explanation:

`cos x` can be considered as a real valued function of real numbers or a complex valued function of complex numbers.

As a real valued function `-1 <= cos x <= 1` for all real values of `x`, so there is no real value of `x` such that `cos x = 3`

To define `cos x` for complex values of `x`, use Euler's formula:

`e^(ix) = cos x + i sin x`

and some basic trigonometric properties:

`cos(-x) = cos(x)`

`sin(-x) = - sin(x)`

Then we find:

`(e^(ix)+e^(-ix))/2 = 1/2((cos(x) + i sin(x))+(cos(-x)+i sin(-x)))`

`color(white)((e^(ix)+e^(-ix))/2) = 1/2((cos(x) + i sin(x))+(cos(x)-i sin(x)))`

`color(white)((e^(ix)+e^(-ix))/2) = cos(x)`

Then we can use the definition:

`cos(x) = 1/2(e^(ix)+e^(-ix))`

to extend the definition of `cos(x)` to complex numbers.

Then to solve `cos(x) = 3`, proceed as follows:

`3 = cos(x) = 1/2(e^(ix)+e^(-ix))`

Multiply both ends by `e^(ix)` and rearranging a little:

`0 = (e^(ix))^2-6(e^(ix))+1`

`color(white)(0) = (e^(ix))^2-6(e^(ix))+9-8`

`color(white)(0) = (e^(ix)-3)^2-(2sqrt(2))^2`

`color(white)(0) = ((e^(ix)-3)-2sqrt(2))((e^(ix)-3)+2sqrt(2))`

`color(white)(0) = (e^(ix)-3-2sqrt(2))(e^(ix)-3+2sqrt(2))`

So:

`e^(ix) = 3+-2sqrt(2)`

So:

`ix = ln(3+-2sqrt(2))+2npii" "` for any `n in ZZ`

Noting that `3-2sqrt(2) = 1/(3+2sqrt(2))`, this becomes:

`x = 2npi+-iln(3+2sqrt(2))" "` for any `n in ZZ`