A.\[x = - \frac{\pi }{4} + k\pi ;\,\,x = \frac{\pi }{6} + k2\pi ;x = \frac{{5\pi }}{6} + k2\pi ;\,\,x = - \frac{\pi }{2} + k2\pi \left( {k \in \mathbb{Z}} \right)\]
B. \[x = \frac{\pi }{4} + k2\pi ;\,\,x = - \frac{\pi }{6} + k2\pi ;x = \frac{{5\pi }}{6} + k\pi ;\,\,x = - \frac{\pi }{2} + k\pi \left( {k \in \mathbb{Z}} \right)\]
C. \[x = \pm \frac{\pi }{6} + k2\pi ;x = \frac{{5\pi }}{6} + k2\pi ;\,\,x = - \frac{\pi }{2} + k2\pi \left( {k \in \mathbb{Z}} \right)\]
D. \[x = - \frac{\pi }{8} + k\pi ;\,\,x = \frac{\pi }{6} + k\pi ;x = - \frac{{5\pi }}{6} + \frac{{k\pi }}{6};\,\,x = - \frac{\pi }{2} + \frac{{k\pi }}{6}\left( {k \in \mathbb{Z}} \right)\]
\[\cos x\cos \frac{x}{2}\cos \frac{{3x}}{2} - \sin x\sin \frac{x}{2}\sin \frac{{3x}}{2} = \frac{1}{2}\]
\[ \Leftrightarrow \frac{1}{2}cosx(cos2x + cosx) + \frac{1}{2}sinx(cos2x - cosx) = \frac{1}{2}\]
\[ \Leftrightarrow cosxcos2x + co{s^2}x + sinxcos2x - sinxcosx = 1\]
\[ \Leftrightarrow cos2x(sinx + cosx) - sinxcosx + co{s^2}x - 1 = 0\]
\[ \Leftrightarrow cos2x(sinx + cosx) - sinxcosx - si{n^2}x = 0\]
\[ \Leftrightarrow cos2x(sinx + cosx) - sinx(sinx + cosx) = 0\]
\[ \Leftrightarrow (sinx + cosx)(cos2x - sinx) = 0\]
\( \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{sinx + cosx = 0}\\{cos2x - sinx = 0}\end{array}} \right.\)
\( \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{sinx = - cosx}\\{1 - 2si{n^2}x - sinx = 0}\end{array}} \right.\)
\( \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{tanx = - 1}\\{sinx = \frac{1}{2}}\\{sinx = - 1}\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = - \frac{\pi }{4} + k\pi }\\{x = \frac{\pi }{6} + k2\pi }\\{x = \frac{{5\pi }}{6} + k2\pi }\\{x = - \frac{\pi }{2} + k2\pi }\end{array}} \right.(k \in \mathbb{Z})\)
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
Câu hỏi trên thuộc đề trắc nghiệm dưới đây !
Copyright © 2021 HOCTAP247