You need to test if the function `cos(ln x)` is invertible in `(0, e^pi]` , hence, you need to perform the following steps, such that:

`y = cos(ln x) => cos^(-1) y = cos^(-1) cos(ln x) cos^(-1) y = ln x => x = e^(cos^(-1) y) `

Since `x > 0` , hence, you may consider `x = 1` , such that:

`1 = e^(cos^(-1) y) => e^0 = e^(cos^(-1) y) => 0 = cos^(-1) y => y = 1`

Considering` x = e^p` i yields:

`e^pi = e^(cos^(-1) y) => pi = cos^(-1) y => y = -1`

**Hence, testing if the function `y = cos(ln x)` is invertible in `(0, e^pi]` , yields that it is, **`f^(-1)(x) = e^(cos^(-1) x).`