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