Xét \[y = f\left( x \right) = \cos \left( {2x - \frac{\pi }{3}} \right)\] Phương trình \[{f^{\left( 4 \right)}}\left( x \right) = - 8\;\]có nghiệm \[x \in \left[ {0;\frac{\pi }{2}} \right]\;\] là:

\[\begin{array}{l}f\prime (x) = - 2sin(2x - \frac{\pi }{3})\\f\prime \prime (x) = - 4cos(2x - \frac{\pi }{3})\\f\prime \prime \prime (x) = 8sin(2x - \frac{\pi }{3})\\{f^{(4)}}(x) = 16cos(2x - \frac{\pi }{3})\\{f^{(4)}}(x) = - 8 \Leftrightarrow cos(2x - \frac{\pi }{3}) = - \frac{1}{2}\end{array}\]

\( \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{2x - \frac{\pi }{3} = \frac{{2\pi }}{3} + k2\pi }\\{2x - \frac{\pi }{3} = - \frac{{2\pi }}{3} + k2\pi }\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = \frac{\pi }{2} + k\pi }\\{x = - \frac{\pi }{6} + k\pi }\end{array}} \right.(k \in Z)\)

\[x \in [0;\frac{\pi }{2}] \Rightarrow x = \frac{\pi }{2}\]

Đáp án cần chọn là: A