Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\sin 4x - \cos 4x = - \sqrt 2 \\
\Leftrightarrow \dfrac{1}{{\sqrt 2 }}.\sin 4x - \dfrac{1}{{\sqrt 2 }}.\cos 4x = - 1\\
\Leftrightarrow \sin 4x.\cos \dfrac{\pi }{4} - \cos 4x.\sin \dfrac{\pi }{4} = - 1\\
\Leftrightarrow \sin \left( {4x - \dfrac{\pi }{4}} \right) = - 1\\
\Leftrightarrow 4x - \dfrac{\pi }{4} = - \dfrac{\pi }{2} + k2\pi \\
\Leftrightarrow 4x = - \dfrac{\pi }{4} + k2\pi \\
\Leftrightarrow x = - \dfrac{\pi }{{16}} + \dfrac{{k\pi }}{2}\,\,\,\,\,\,\,\,\,\left( {k \in Z} \right)\\
b,\\
\sin 3x + \cos 3x = \sqrt 2 \\
\Leftrightarrow \dfrac{1}{{\sqrt 2 }}.\sin 3x + \dfrac{1}{{\sqrt 2 }}.\cos 3x = 1\\
\Leftrightarrow \sin 3x.\cos \dfrac{\pi }{4} + \cos 3x.\sin \dfrac{\pi }{4} = 1\\
\Leftrightarrow \sin \left( {3x + \dfrac{\pi }{4}} \right) = 1\\
\Leftrightarrow 3x + \dfrac{\pi }{4} = \dfrac{\pi }{2} + k2\pi \\
\Leftrightarrow 3x = \dfrac{\pi }{4} + k2\pi \\
\Leftrightarrow x = \dfrac{\pi }{{12}} + \dfrac{{k2\pi }}{3}\,\,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
