Đáp án:
\[x = \dfrac{{3\pi }}{2} + k4\pi \,\,\,\,\left( {k \in Z} \right)\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\sin \dfrac{x}{2} - \cos \dfrac{x}{2} = \sqrt 2 \\
\Leftrightarrow \dfrac{{\sqrt 2 }}{2}.\sin \dfrac{x}{2} - \dfrac{{\sqrt 2 }}{2}.\cos \dfrac{x}{2} = \dfrac{{\sqrt 2 }}{2}.\sqrt 2 \\
\Leftrightarrow \sin \dfrac{x}{2}.\cos \dfrac{\pi }{4} - \cos \dfrac{x}{2}.\sin \dfrac{\pi }{4} = 1\\
\Leftrightarrow \sin \left( {\dfrac{x}{2} - \dfrac{\pi }{4}} \right) = 1\\
\Leftrightarrow \dfrac{x}{2} - \dfrac{\pi }{4} = \dfrac{\pi }{2} + k2\pi \\
\Leftrightarrow \dfrac{x}{2} = \dfrac{{3\pi }}{4} + k2\pi \\
\Leftrightarrow x = \dfrac{{3\pi }}{2} + k4\pi \,\,\,\,\left( {k \in Z} \right)
\end{array}\)
