Đáp án:1a/ TXĐ:
$D = R\backslash \left\{ {\frac{\pi }{6} + \frac{{k\pi }}{3},k \in Z} \right\}$
b/
$[_{x = \frac{\pi }{4} + k\pi }^{x = \arctan 7 + k\pi },k \in Z$
Giải thích các bước giải:a/
$\begin{array}{l}
ĐKXĐ :\cos 3x \ne 0 < = > 3x \ne \frac{\pi }{2} + k\pi < = > x \ne \frac{\pi }{6} + \frac{{k\pi }}{3}\\
TXĐ: D = R\backslash \left\{ {\frac{\pi }{6} + \frac{{k\pi }}{3},k \in Z} \right\}
\end{array}$
b/
$\begin{array}{l}
{\sin ^2}x - 8\sin x\cos x + 7{\cos ^2}x = 0\\
+ \cos x = 0\\
{\sin ^2}x + {\cos ^2}x = 1\\
= > {\sin ^2}x = 1\\
PTTT:{\sin ^2}x = 0(vô lý)\\
+ \cos x \ne 0\\
chia 2 vế cho{\cos ^2}x\\
PTTT:{\left( {\frac{{\sin x}}{{\cos x}}} \right)^2} - 8\frac{{\sin x}}{{\cos x}} + 7 = 0\\
< = > {\tan ^2}x - 8\tan x + 7 = 0\\
< = > [_{\tan x = 1 = \tan \frac{\pi }{4}}^{\tan x = 7}\\
< = > [_{x = \frac{\pi }{4} + k\pi }^{x = \arctan 7 + k\pi },k \in Z
\end{array}$
