Official solution
Let z = \[{{3 + i\sin \theta } \over {4 - i\cos \theta }}\]
= \[{{3 + i\sin \theta } \over {4 - i\cos \theta }} \times {{\left( {4 + i\cos \theta } \right)} \over {\left( {4 + i\cos \theta } \right)}}\]
= \[{{\left( {12 - \sin \theta \cos \theta } \right) + i\left( {4\sin \theta + 3\cos \theta } \right)} \over {16 + {{\cos }^2}\theta }}\]
As Z is purely real
\[ \therefore \] 4sin\[\theta \] + 3cos\[\theta \] = 0
\[ \Rightarrow \] tan \[\theta \] = \[ - {3 \over 4}\]
\[ \therefore \] \[\theta \] lies in the 2nd quadrant then
arg(sin\theta + icos\theta ) = \[\pi \] + \[{\tan ^{ - 1}}\left( {{{\cos \theta } \over {\sin \theta }}} \right)\]
= \[\pi - {\tan ^{ - 1}}\left( {{4 \over 3}} \right)\]