Đáp án: 1.A    2.C

Giải thích các bước giải:

$\begin{array}{l}
1)\\
8co{s^3}\left( {x + \frac{\pi }{3}} \right) = \cos 3x\\
 \Rightarrow 8co{s^3}\left( {x + \frac{\pi }{3}} \right) =  - \cos \left( {3x + \pi } \right)\\
 \Rightarrow 8co{s^3}\left( {x + \frac{\pi }{3}} \right) =  - \cos \left[ {3\left( {x + \frac{\pi }{3}} \right)} \right]\\
 \Rightarrow 8co{s^3}\left( {x + \frac{\pi }{3}} \right) =  - \left( {4{{\cos }^3}\left( {x + \frac{\pi }{3}} \right) - 3\cos \left( {x + \frac{\pi }{3}} \right)} \right)\\
 \Rightarrow 12{\cos ^3}\left( {x + \frac{\pi }{3}} \right) - 3\cos \left( {x + \frac{\pi }{3}} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
\cos \left( {x + \frac{\pi }{3}} \right) = 0\\
{\cos ^2}\left( {x + \frac{\pi }{3}} \right) = \frac{1}{4}
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\cos \left( {x + \frac{\pi }{3}} \right) = 0\\
\cos \left( {x + \frac{\pi }{3}} \right) = \frac{1}{2}\\
\cos \left( {x + \frac{\pi }{3}} \right) =  - \frac{1}{2}
\end{array} \right.\\
 \Rightarrow 6\,điểm\,biểu\,diễn\\
2)\frac{{\tan 2x}}{{\sin x - \cos x}} = \frac{{\tan 2x}}{{\sqrt 2 \sin \left( {x - \frac{\pi }{4}} \right)}}\\
Xác định:\left\{ \begin{array}{l}
\cos 2x \ne 0\\
\sin \left( {x - \frac{\pi }{4}} \right) \ne 0
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
2x \ne \frac{\pi }{2} + k\pi \\
x - \frac{\pi }{4} \ne k\pi 
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x \ne \frac{\pi }{4} + \frac{{k\pi }}{2}\\
x \ne \frac{\pi }{4} + k\pi 
\end{array} \right.\\
 \Rightarrow x \ne \frac{\pi }{4} + \frac{{k\pi }}{2}\\
 \Rightarrow C
\end{array}$