Official solution
\[f\left( x \right) = {4^{ - {x^2}}} + {\cos ^{ - 1}}\left( {{x \over 2} - 1} \right) + \log \left( {\cos \,x} \right)\]
\[f\left( x \right)\] is defined if \[ - 1 \le \left( {{x \over 2} - 1} \right) \le 1\] and \[\cos \,x > 0\]
or \[\,\,\,\,0 \le {x \over 2} \le 2\,\,\] and \[\,\, - {\pi \over 2} < x < {\pi \over 2}\]
or \[\,\,\,\] \[0 \le x \le 4\] \[\,\,\] and \[\,\, - {\pi \over 2} < x < {\pi \over 2}\]
\[\therefore\] \[\,\,\,\,\,x\, \in \left[ {0,{\pi \over 2}} \right)\]