Giải phương trình √3 sin3x – cos3x = √2.
\(\eqalign{
& \sqrt 3 \sin 3x – \cos 3x = \sqrt 2 \cr
& \Leftrightarrow {{\sqrt 3 } \over 2}\sin 3x – {1 \over 2}\cos 3x = {{\sqrt 2 } \over 2} \cr
& \Leftrightarrow \cos {\pi \over 6}\sin 3x – \sin {\pi \over 6}\cos 3x = \sin {\pi \over 4} \cr
& \Leftrightarrow \sin (3x – {\pi \over 6}) = \sin {\pi \over 6} \cr
& \Leftrightarrow \left[ \matrix{
3x – {\pi \over 6} = {\pi \over 6} + k2\pi \hfill \cr
3x – {\pi \over 6} = \pi – {\pi \over 6}\pi + k2\pi \hfill \cr} \right.;\,k \in Z \cr
& \Leftrightarrow \left[ \matrix{
3x = {\pi \over 6} + k2\pi \hfill \cr
3x = \pi + k2\pi \hfill \cr} \right.;k \in Z \cr
& \Leftrightarrow \left[ \matrix{
x = {\pi \over 9} + k{{2\pi } \over 3} \hfill \cr
x = {\pi \over 3} + k{{2\pi } \over 3} \hfill \cr} \right.;\,\,k \in Z \cr} \)