Đáp án:
\(\left[ \matrix{
x = {\pi \over {12}} + k\pi \hfill \cr
x = {{5\pi } \over {12}} + k\pi \hfill \cr} \right.\,\,\left( {k \in Z} \right)\)
Giải thích các bước giải:
\(\eqalign{
& 3\tan 2x - 2\sin \left( {2x - {{3\pi } \over 2}} \right) + {{2\left( {\cos x - \sin x} \right)} \over {\cos x + \sin x}} = {1 \over {\cos 2x}}\,\,\left( {\cos 2x \ne 0} \right) \cr
& \Leftrightarrow 3\tan 2x - 2\cos 2x + {{2\left( {\cos x - \sin x} \right)} \over {\cos x + \sin x}} = {1 \over {\cos 2x}} \cr
& \Leftrightarrow {{3\sin 2x} \over {\cos 2x}} - {{2{{\cos }^2}2x} \over {\cos 2x}} + {{2{{\left( {\cos x - \sin x} \right)}^2}} \over {\cos 2x}} = {1 \over {\cos 2x}} \cr
& \Leftrightarrow 3\sin 2x - 2{\cos ^2}2x + 2\left( {1 - \sin 2x} \right) = 1 \cr
& \Leftrightarrow 3\sin 2x - 2\left( {1 - {{\sin }^2}2x} \right) + 2 - 2\sin 2x = 1 \cr
& \Leftrightarrow 3\sin 2x - 2 + 2{\sin ^2}2x + 2 - 2\sin 2x = 1 \cr
& \Leftrightarrow 2{\sin ^2}2x + \sin 2x - 1 = 0 \cr
& \Leftrightarrow \left[ \matrix{
\sin 2x = - 1\,\,\left( {loai} \right) \hfill \cr
\sin 2x = {1 \over 2} \hfill \cr} \right. \cr
& \Leftrightarrow \left[ \matrix{
2x = {\pi \over 6} + k2\pi \hfill \cr
2x = {{5\pi } \over 6} + k2\pi \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
x = {\pi \over {12}} + k\pi \hfill \cr
x = {{5\pi } \over {12}} + k\pi \hfill \cr} \right.\,\,\left( {k \in Z} \right) \cr} \)
