Đáp án:
\(\left\{ \matrix{ x = - {\pi \over 6} + k2\pi \hfill \cr x = {\pi \over 2} + k2\pi \hfill \cr} \right.\,\,\left( {k \in Z} \right)\)
Giải thích các bước giải:
\(\eqalign{ & \cos 2\left( {x + {\pi \over 3}} \right) + 4\cos \left( {{\pi \over 6} - x} \right) = {5 \over 2} \cr & \Leftrightarrow \cos 2\left( {x + {\pi \over 3}} \right) + 4\sin \left( {x + {\pi \over 3}} \right) = {5 \over 2} \cr & \Leftrightarrow 1 - 2{\sin ^2}\left( {x + {\pi \over 3}} \right) + 4\sin \left( {x + {\pi \over 3}} \right) = {5 \over 2} \cr & \Leftrightarrow - 2{\sin ^2}\left( {x + {\pi \over 3}} \right) + 4\sin \left( {x + {\pi \over 3}} \right) - {3 \over 2} = 0 \cr & \Leftrightarrow \left[ \matrix{ \sin \left( {x + {\pi \over 3}} \right) = {3 \over 2}\,\,\left( {loai} \right) \hfill \cr \sin \left( {x + {\pi \over 3}} \right) = {1 \over 2}\,\,\left( {tm} \right) \hfill \cr} \right. \cr & \Leftrightarrow \left[ \matrix{ x + {\pi \over 3} = {\pi \over 6} + k2\pi \hfill \cr x + {\pi \over 3} = {{5\pi } \over 6} + k2\pi \hfill \cr} \right. \cr & \Leftrightarrow \left[ \matrix{ x = - {\pi \over 6} + k2\pi \hfill \cr x = {\pi \over 2} + k2\pi \hfill \cr} \right.\,\,\left( {k \in Z} \right) \cr} \)
