Đáp án:
\[\left[ \begin{array}{l}
x = \dfrac{\pi }{{12}} + \dfrac{{k2\pi }}{3}\\
x = \dfrac{{3\pi }}{4} + k2\pi
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\cos x - \sin x = \sqrt 2 .\sin 2x\\
\Leftrightarrow \dfrac{{\sqrt 2 }}{2}.\cos x - \dfrac{{\sqrt 2 }}{2}\sin x = \dfrac{{\sqrt 2 }}{2}.\sqrt 2 \sin 2x\\
\Leftrightarrow \cos x.\cos \dfrac{\pi }{4} - \sin x.\sin \dfrac{\pi }{4} = \sin 2x\\
\Leftrightarrow \cos \left( {x + \dfrac{\pi }{4}} \right) = \sin 2x\\
\Leftrightarrow \cos \left( {x + \dfrac{\pi }{4}} \right) = \cos \left( {\dfrac{\pi }{2} - 2x} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{\pi }{4} = \dfrac{\pi }{2} - 2x + k2\pi \\
x + \dfrac{\pi }{4} = 2x - \dfrac{\pi }{2} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
3x = \dfrac{\pi }{4} + k2\pi \\
- x = - \dfrac{{3\pi }}{4} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{{12}} + \dfrac{{k2\pi }}{3}\\
x = \dfrac{{3\pi }}{4} + k2\pi
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
