Hey Eraser147.
Try undoing the transformations. You can apply an inverse cosine to get the possible solutions (and use the fact that cos is periodic).
So then you have e^x + 1 = n*2*pi + pi/2 fo integer values n.
Now x has to be minimized for this to hold and e^x is unique and > 0 which means that you can't have negative values for n so n = 0. This means e^x + 1 = pi/2 since pi/2 > 1 but less than all the others (the minimal branch).
So now you have e^x = pi/2 - 1. and taking logarithms gives you x = ln(pi/2 - 1).
The given equation implies:
$\displaystyle e^x+1.0=\frac{\pi}{2}+k\pi$
Since $\displaystyle 0<1<\frac{\pi}{2}$ we want $\displaystyle k=0$ and so:
$\displaystyle e^x+1.0=\frac{\pi}{2}$
$\displaystyle e^x=\frac{\pi}{2}-1.0$
Convert from exponential lo logarithmic form:
$\displaystyle x=\ln\left(\frac{\pi}{2}-1.0 \right)$