Đáp án:
\(x = {{k\pi } \over 2}\,\left( {k \in Z} \right)\)
Giải thích các bước giải:
\(\eqalign{
& \left| {\sin x - \cos x} \right| + 4\sin 2x = 1 \cr
& Dat\,\,t = \left| {\sin x - \cos x} \right|\,\,\left( {0 \le t \le \sqrt 2 } \right) \cr
& \Rightarrow {t^2} = 1 - 2\sin x\cos x \Leftrightarrow \sin 2x = 1 - {t^2} \cr
& \Rightarrow t + 4\left( {1 - {t^2}} \right) = 1 \cr
& \Leftrightarrow t + 4 - 4{t^2} = 1 \Leftrightarrow 4{t^2} - t - 3 = 0 \cr
& \Leftrightarrow \left[ \matrix{
t = 1\,\,\,\left( {tm} \right) \hfill \cr
t = - {3 \over 4}\,\,\left( {loai} \right) \hfill \cr} \right. \cr
& \Leftrightarrow \left| {\sin x - \cos x} \right| = 1 \cr
& \Leftrightarrow \sin 2x = 1 - 1 = 0 \Leftrightarrow 2x = k\pi \Leftrightarrow x = {{k\pi } \over 2}\,\left( {k \in Z} \right) \cr} \)
