Xét phương trình: (2cos2x + 5)(sin⁴x – cos⁴x) + 3 = 0

⇔ (2cos2x + 5)(sin²x – cos²x) + 3 = 0

⇔ - (2cos2x + 5)cos2x + 3 = 0

⇔ - 2cos²2x - 5cos2x + 3 = 0

\( \Leftrightarrow \left[ \begin{array}{l}{\rm{cos}}2x = \frac{1}{2}\\\cos 2x = - 3(L)\end{array} \right.\)

\( \Leftrightarrow {\rm{cos}}2x = \frac{1}{2}\)

\( \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{6} + k\pi \\x = - \frac{\pi }{6} + k\pi \end{array} \right.,k \in \mathbb{Z}\)

Vì \(x \in \left( {0;2\pi } \right)\) nên \(\left[ \begin{array}{l}0 < \frac{\pi }{6} + k\pi < 2\pi \\0 < - \frac{\pi }{6} + k\pi < 2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} - \frac{1}{6} < k < \frac{{11}}{6}\\\frac{1}{6} < k < \frac{{13}}{6}\end{array} \right.\)

Mà \(k \in \mathbb{Z}\)

\(\left[ \begin{array}{l}k \in \left\{ {0;1} \right\} \Rightarrow x = \frac{\pi }{6},x = \frac{{7\pi }}{6}\\k \in \left\{ {1;2} \right\} \Rightarrow x = \frac{{5\pi }}{6},x = \frac{{11\pi }}{6}\end{array} \right.\)

\( \Rightarrow x \in \left\{ {\frac{\pi }{6},\frac{{5\pi }}{6},\frac{{7\pi }}{6};\frac{{11\pi }}{6}} \right\}\)

Vậy tổng các nghiệm \(S = \frac{\pi }{6} + \frac{{5\pi }}{6} + \frac{{7\pi }}{6} + \frac{{11\pi }}{6} = 4\pi \).

Chọn B.

Số nghiệm của phương trình 2sin^2x+cos2x+1=0 trong đoạn 0;2018pi