Đáp án:
\(\left[ \begin{array}{l}
x = \dfrac{{\dfrac{\pi }{4} - \arcsin \dfrac{1}{{3\sqrt 2 }} + k2\pi }}{2}\\
x = \dfrac{{ - \dfrac{{3\pi }}{4} + \arcsin \dfrac{1}{{3\sqrt 2 }} + k2\pi }}{2}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
\cos 2x - \sin 2x = \dfrac{1}{3}\\
\to \dfrac{1}{{\sqrt 2 }}\cos 2x - \dfrac{1}{{\sqrt 2 }}.\sin 2x = \dfrac{1}{{3\sqrt 2 }}\\
\to \sin \dfrac{\pi }{4}.\cos 2x - \cos \dfrac{\pi }{4}.\sin 2x = \dfrac{1}{{3\sqrt 2 }}\\
\to \sin \left( {\dfrac{\pi }{4} - 2x} \right) = \dfrac{1}{{3\sqrt 2 }}\\
\to \left[ \begin{array}{l}
\dfrac{\pi }{4} - 2x = \arcsin \dfrac{1}{{3\sqrt 2 }} + k2\pi \\
\dfrac{\pi }{4} - 2x = \pi - \arcsin \dfrac{1}{{3\sqrt 2 }} + k2\pi
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{{\dfrac{\pi }{4} - \arcsin \dfrac{1}{{3\sqrt 2 }} + k2\pi }}{2}\\
x = \dfrac{{ - \dfrac{{3\pi }}{4} + \arcsin \dfrac{1}{{3\sqrt 2 }} + k2\pi }}{2}
\end{array} \right.
\end{array}\)
