Đáp án:

\(\begin{array}{l}
1,\\
x = \dfrac{\pi }{4} + \dfrac{{k2\pi }}{3}\,\,\,\,\left( {k \in Z} \right)\\
2,\\
\left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
x = \dfrac{\pi }{{12}} + \dfrac{{k\pi }}{7}
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1,\\
\sin x + \cos x = 2\sqrt 2 \sin x.\cos x\\
 \Leftrightarrow \dfrac{1}{{\sqrt 2 }}\sin x + \dfrac{1}{{\sqrt 2 }}\cos x = 2\sin x.\cos x\\
 \Leftrightarrow \sin x.\cos \dfrac{\pi }{4} + \cos x.\sin \dfrac{\pi }{4} = \sin 2x\\
 \Leftrightarrow \sin \left( {x + \dfrac{\pi }{4}} \right) = 2x\\
 \Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{\pi }{4} = 2x + k2\pi \\
x + \dfrac{\pi }{4} = \pi  - 2x + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 2x =  - \dfrac{\pi }{4} + k2\pi \\
x + 2x = \pi  - \dfrac{\pi }{4} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
 - x =  - \dfrac{\pi }{4} + k2\pi \\
3x = \dfrac{{3\pi }}{4} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k2\pi \\
x = \dfrac{\pi }{4} + \dfrac{{k2\pi }}{3}
\end{array} \right.\\
 \Leftrightarrow x = \dfrac{\pi }{4} + \dfrac{{k2\pi }}{3}\,\,\,\,\left( {k \in Z} \right)\\
2,\\
\sin 8x - \cos 6x = \sqrt 3 \left( {\sin 6x + \cos 8x} \right)\\
 \Leftrightarrow \sin 8x - \sqrt 3 \cos 8x = \sqrt 3 \sin 6x + \cos 6x\\
 \Leftrightarrow \dfrac{1}{2}\sin 8x - \dfrac{{\sqrt 3 }}{2}\cos 8x = \dfrac{{\sqrt 3 }}{2}\sin 6x + \dfrac{1}{2}\cos 6x\\
 \Leftrightarrow \sin 8x.\cos \dfrac{\pi }{3} - \cos 8x.\sin \dfrac{\pi }{3} = \sin 6x.\cos \dfrac{\pi }{6} + \cos 6x.\sin \dfrac{\pi }{6}\\
 \Leftrightarrow \sin \left( {8x - \dfrac{\pi }{3}} \right) = \sin \left( {6x + \dfrac{\pi }{6}} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
8x - \dfrac{\pi }{3} = 6x + \dfrac{\pi }{6} + k2\pi \\
8x - \dfrac{\pi }{3} = \pi  - 6x - \dfrac{\pi }{6} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
8x - 6x = \dfrac{\pi }{6} + \dfrac{\pi }{3} + k2\pi \\
8x + 6x = \pi  - \dfrac{\pi }{6} + \dfrac{\pi }{3} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2x = \dfrac{\pi }{2} + k2\pi \\
14x = \dfrac{{7\pi }}{6} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
x = \dfrac{\pi }{{12}} + \dfrac{{k\pi }}{7}
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)