Đáp án:
\(\left[ \matrix{
x = {\pi \over {12}} + {{k\pi } \over 3} \hfill \cr
x = {1 \over 3}\arctan {1 \over 3} + {{k\pi } \over 3} \hfill \cr} \right.\,\,\left( {k \in Z} \right)\)
Giải thích các bước giải:
\(\eqalign{
& 3\tan 3x + \cot 3x - 4 = 0 \cr
& DK:\,\,3x \ne k\pi \Leftrightarrow x \ne {{k\pi } \over 3}\,\,\left( {k \in Z} \right) \cr
& Pt \Leftrightarrow 3\tan 3x + {1 \over {\tan 3x}} - 4 = 0 \cr
& \Leftrightarrow 3{\tan ^2}3x - 4\tan 3x + 1 = 0 \cr
& \Leftrightarrow \left[ \matrix{
\tan 3x = 1 \hfill \cr
\tan 3x = {1 \over 3} \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
3x = {\pi \over 4}k\pi \hfill \cr
3x = \arctan {1 \over 3} + k\pi \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
x = {\pi \over {12}} + {{k\pi } \over 3} \hfill \cr
x = {1 \over 3}\arctan {1 \over 3} + {{k\pi } \over 3} \hfill \cr} \right.\,\,\left( {k \in Z} \right) \cr} \)
