A. \(x = - \dfrac{\pi }{{12}} - k2\pi ;\,\,x = \dfrac{{19\pi }}{{12}} - k2\pi \)
B. \(x = \dfrac{{7\pi }}{{12}} + k2\pi ;\,\,x = - \dfrac{{13\pi }}{{12}} + k2\pi \)
C. \(x = \dfrac{\pi }{{12}} + k2\pi ;\,\,x = \dfrac{{19\pi }}{{12}} + k2\pi \)
D. \(x = - \dfrac{{7\pi }}{{12}} - k2\pi ;\,\,x = \dfrac{{13\pi }}{{12}} - k2\pi \)
B
Ta có: \(\cos x - \sin x = - \dfrac{{\sqrt 6 }}{2} \) \(\Leftrightarrow \sqrt 2 \cos \left( {x + \dfrac{\pi }{4}} \right) = - \dfrac{{\sqrt 6 }}{2}\)
\( \Leftrightarrow \cos \left( {x + \dfrac{\pi }{4}} \right) = - \dfrac{{\sqrt 3 }}{2}\) \( \Leftrightarrow \cos \left( {x + \dfrac{\pi }{4}} \right) = \cos \dfrac{{5\pi }}{6}\)
\( \Leftrightarrow \left[ \begin{array}{l}x + \dfrac{\pi }{4} = \dfrac{{5\pi }}{6} + k2\pi \\x + \dfrac{\pi }{4} = - \dfrac{{5\pi }}{6} + k2\pi \end{array} \right. \) \(\Leftrightarrow \left[ \begin{array}{l}x = \dfrac{{7\pi }}{{12}} + k2\pi \\x = - \dfrac{{13\pi }}{{12}} + k2\pi \end{array} \right.\;\left( {k \in \mathbb{Z}} \right)\)
Chọn đáp án B.
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