$\begin{array}{l} \sin x + \sin \left( {x + a} \right) + \sin \left( {x + 2a} \right) + \sin \left( {x + 3a} \right) + \sin \left( {x + 4a} \right)\\ = \left( {\sin x + \sin \left( {x + 4a} \right)} \right) + \left( {\sin \left( {x + 3a} \right) + \sin \left( {x + a} \right)} \right) + \sin \left( {x + 2a} \right)\\ = 2\sin \left( {x + 2a} \right)\cos 2a + 2\sin \left( {x + 2a} \right)\cos a + \sin \left( {x + 2a} \right)\\ = \sin \left( {x + 2a} \right)\left( {2\cos 2a + 2\cos a + 1} \right) \end{array}$
Để cho biểu thức không phụ thuộc vào $x$ thì
$2\cos2a+2\cos a+1=0$
Vì
$\begin{array}{l} a \in \left( {0;\pi } \right) \Rightarrow \cos a - 1 \ne 0\\ 2\cos 2a + 2\cos a + 1 = 0\\ \Leftrightarrow 2\left( {2{{\cos }^2}a - 1} \right) + 2\cos a + 1 = 0\\ \Leftrightarrow 4{\cos ^2}a + 2\cos a - 1 = 0\\ \Leftrightarrow \left( {\cos a - 1} \right)\left( {4{{\cos }^2}a + 2\cos a - 1} \right) = 0\\ \Leftrightarrow 4{\cos ^3}a + 2{\cos ^2}a - \cos a - 4{\cos ^2}a - 2\cos a + 1 = 0\\ \Leftrightarrow 4{\cos ^3}a - 3\cos a = 2{\cos ^2}a - 1\\ \Leftrightarrow \cos 3a = \cos 2a\\ \Leftrightarrow \cos \left( { - 3a} \right) = \cos 2a\\ \Leftrightarrow \cos \left( {2\pi - 3a} \right) = \cos 2a\\ \Leftrightarrow 2\pi - 3a = 2a\\ \Leftrightarrow 2\pi = 5a \Leftrightarrow a = \dfrac{{2\pi }}{5} \end{array}$
