Official solution
\[(\sin x + \sin 4x) + (\sin 2x + \sin 3x) = 0\]
\[ \Rightarrow 2\sin {{5x} \over 2}\left\{ {\cos {{3x} \over 2} + \cos {x \over 2}} \right\} = 0\]
\[ \Rightarrow 2\sin {{5x} \over 2}\left\{ {2\cos x\cos {x \over 2}} \right\} = 0\]
\[2\sin {{5x} \over 2} = 0 \Rightarrow {{5x} \over 2} = 0,\pi ,2\pi ,3\pi ,4\pi ,5\pi \]
\[ \Rightarrow x = 0,{{2\pi } \over 5},{{4\pi } \over 5},{{6\pi } \over 5},{{8\pi } \over 5},2\pi \]
\[\cos {x \over 2} = 0 \Rightarrow {x \over 2} = {\pi \over 2} \Rightarrow x = \pi \]
\[\cos x = 0 \Rightarrow x = {\pi \over 2},{{3\pi } \over 2}\]
So, sum \[ = 6\pi + \pi + 2\pi = 9\pi \]