\(\sin \left( {2x + \frac{{9\pi }}{2}} \right) - 3\cos \left( {x - \frac{{15\pi }}{2}} \right) = 1 + 2\sin x\)

\( \Leftrightarrow \sin \left( {2x + \frac{\pi }{2} + 4\pi } \right) - 3\cos \left( {x + \frac{\pi }{2} - 8\pi } \right) = 1 + 2\sin x\)

\( \Leftrightarrow \sin \left( {2x + \frac{\pi }{2}} \right) - 3\cos \left( {x + \frac{\pi }{2}} \right) = 1 + 2\sin x\)

\( \Leftrightarrow \cos 2x + 3\sin x - 2\sin x - 1 = 0\)

\( \Leftrightarrow 1 - 2{\sin ^2}x + \sin x - 1 = 0\)

\( \Leftrightarrow \sin x(2\sin x - 1) = 0\)

\( \Leftrightarrow \left[ {\begin{array}{*{20}{l}}{\sin x = 0}\\{\sin x = \frac{1}{2}}\end{array} \Leftrightarrow \left[ {\begin{array}{*{20}{l}}{x = k\pi }\\{x = \frac{\pi }{6} + k2\pi }\\{x = \frac{{5\pi }}{6} + k2\pi }\end{array}\quad (k \in \mathbb{Z})} \right.} \right.\)

Các nghiệm thuộc đoạn \([0;2\pi ]\) là \(0,\pi ,2\pi ,\frac{\pi }{6},\frac{{5\pi }}{6}\).

Vậy \(S = 0 + \pi + 2\pi + \frac{\pi }{6} + \frac{{5\pi }}{6} = 4\pi \). Chọn A.