\(\left(2cos2x+5\right)\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)+3=0\)
\(\Leftrightarrow-cos2x\left(2cos2x+5\right)+3=0\)
Đặt \(cos2x=a\) (\(-1\le a< 1\))
\(\Leftrightarrow2a^2+5a-3=0\) \(\Rightarrow\left[{}\begin{matrix}a=\frac{1}{2}\\a=-3< -1\left(l\right)\end{matrix}\right.\)
\(\Rightarrow cos2x=\frac{1}{2}\Rightarrow\left[{}\begin{matrix}2x=\frac{\pi}{3}+k2\pi\\2x=-\frac{\pi}{3}+l2\pi\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{6}+k\pi\\x=-\frac{\pi}{6}+l\pi\end{matrix}\right.\)
Do \(x\in\left(0;2\pi\right)\Rightarrow x=\left\{\frac{\pi}{6};\frac{5\pi}{6};\frac{11\pi}{6}\right\}\) \(\Rightarrow\sum x=\frac{17\pi}{6}\)
