Giải phương trình \[\cos x\cos \frac{x}{2}\cos \frac{2} - \sin x\sin \frac{x}{2}\sin \frac{2} = \frac{1}{2}\]

\[\cos x\cos \frac{x}{2}\cos \frac{2} - \sin x\sin \frac{x}{2}\sin \frac{2} = \frac{1}{2}\]

\[ \Leftrightarrow \frac{1}{2}cosx(cos2x + cosx) + \frac{1}{2}sinx(cos2x - cosx) = \frac{1}{2}\]

\[ \Leftrightarrow cosxcos2x + co{s^2}x + sinxcos2x - sinxcosx = 1\]

\[ \Leftrightarrow cos2x(sinx + cosx) - sinxcosx + co{s^2}x - 1 = 0\]

\[ \Leftrightarrow cos2x(sinx + cosx) - sinxcosx - si{n^2}x = 0\]

\[ \Leftrightarrow cos2x(sinx + cosx) - sinx(sinx + cosx) = 0\]

\[ \Leftrightarrow (sinx + cosx)(cos2x - sinx) = 0\]

\( \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{sinx + cosx = 0}\\{cos2x - sinx = 0}\end{array}} \right.\)

\( \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{sinx = - cosx}\\{1 - 2si{n^2}x - sinx = 0}\end{array}} \right.\)

\( \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{tanx = - 1}\\{sinx = \frac{1}{2}}\\{sinx = - 1}\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = - \frac{\pi }{4} + k\pi }\\{x = \frac{\pi }{6} + k2\pi }\\{x = \frac{{5\pi }}{6} + k2\pi }\\{x = - \frac{\pi }{2} + k2\pi }\end{array}} \right.(k \in \mathbb{Z})\)

Vậy nghiệm của phương trình đã cho là: \[x = - \frac{\pi }{4} + k\pi ;\,\,x = \frac{\pi }{6} + k2\pi ;x = \frac{{5\pi }}{6} + k2\pi ;\,\,x = - \frac{\pi }{2} + k2\pi \left( {k \in \mathbb{Z}} \right)\]

Đáp án cần chọn là: A