Đáp án:
\[x = - \dfrac{{7\pi }}{{24}} + k\pi \,\,\,\,\left( {k \in Z} \right)\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\sin x + \cos x + \sqrt 2 \cos \left( {\dfrac{\pi }{6} - x} \right) = 0\\
\Leftrightarrow \sin x + \cos x = - \sqrt 2 \cos \left( {\dfrac{\pi }{6} - x} \right)\\
\Leftrightarrow \dfrac{{\sqrt 2 }}{2}\cos x + \dfrac{{\sqrt 2 }}{2}.\sin x = - \dfrac{{\sqrt 2 }}{2}.\sqrt 2 \cos \left( {\dfrac{\pi }{6} - x} \right)\\
\Leftrightarrow \cos x.\cos \dfrac{\pi }{4} + \sin x.\sin \dfrac{\pi }{4} = - \cos \left( {\dfrac{\pi }{6} - x} \right)\\
\Leftrightarrow \cos \left( {x - \dfrac{\pi }{4}} \right) = \cos \left[ {\pi - \left( {\dfrac{\pi }{6} - x} \right)} \right]\\
\Leftrightarrow \cos \left( {x - \dfrac{\pi }{4}} \right) = \cos \left( {\dfrac{{5\pi }}{6} + x} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x - \dfrac{\pi }{4} = \dfrac{{5\pi }}{6} + x + k2\pi \\
x - \dfrac{\pi }{4} = - \dfrac{{5\pi }}{6} - x + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
- \dfrac{{13\pi }}{{12}} = k2\pi \,\,\,\,\left( L \right)\\
2x = - \dfrac{{7\pi }}{{12}} + k2\pi
\end{array} \right.\\
\Leftrightarrow x = - \dfrac{{7\pi }}{{24}} + k\pi \,\,\,\,\left( {k \in Z} \right)
\end{array}\)
