Đáp án:
\(\left[ \matrix{
x = {\pi \over 4} + k\pi \hfill \cr
x = \arctan \left( { - {1 \over 3}} \right) + k\pi \hfill \cr} \right.\,\,\left( {k \in Z} \right)\)
Giải thích các bước giải:
$$\eqalign{
& 3{\sin ^2}x - 2\sin x\cos x - {\cos ^2}x = 0 \cr
& TH1:\,\,\cos x = 0 \Leftrightarrow {\sin ^2}x = 1 \Leftrightarrow x = {\pi \over 2} + k\pi \,\,\left( {k \in Z} \right) \cr
& \Rightarrow 3.1 - 2.0 - 0 = 0\,\,\left( {Vo\,\,nghiem} \right) \cr
& TH2:\,\,\cos x \ne 0 \Leftrightarrow x \ne {\pi \over 2} + k\pi \cr
& PT \Leftrightarrow 3{\tan ^2}x - 2\tan x - 1 = 0 \cr
& \Leftrightarrow \left[ \matrix{
\tan x = 1 \hfill \cr
\tan x = - {1 \over 3} \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
x = {\pi \over 4} + k\pi \hfill \cr
x = \arctan \left( { - {1 \over 3}} \right) + k\pi \hfill \cr} \right.\,\,\left( {k \in Z} \right) \cr} $$
