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Giải các phương trình sau :
a. \(y’ = 0\,voi\,y = {1 \over 2}\sin 2x + \sin x – 3\)
b. \(y’ = 0,\,voi\,y = \sin 3x – 2\cos 3x – 3x + 4\)
a. Ta có:
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\(\eqalign{ & y’ = \cos 2x + \cos x \cr & y’ = 0 \Leftrightarrow \cos 2x + \cos x = 0 \cr & \Leftrightarrow 2{\cos ^2}x + \cos x – 1 = 0 \cr & \Leftrightarrow \left[ {\matrix{ {\cos x = – 1} \cr {\cos x = {1 \over 2}} \cr } } \right. \Leftrightarrow \left[ {\matrix{ {x = \pi + k2\pi } \cr {x = \pm {\pi \over 3} + k2\pi } \cr } } \right.\,\,\left( {k \in Z} \right) \cr} \)
b.
\(\eqalign{ & y’ = 3\cos 3x + 6\sin 3x – 3 \cr & y’ = 0 \Leftrightarrow \cos 3x + 2\sin 3x = 1 \cr & \Leftrightarrow {1 \over {\sqrt 5 }}\cos 3x + {2 \over {\sqrt 5 }}\sin 3x = {1 \over {\sqrt 5 }} \cr & \Leftrightarrow \cos \left( {3x – \alpha } \right) = \cos \alpha \,\left( {voi\,\cos \alpha = {1 \over {\sqrt 5 }}} \right) \cr & \Leftrightarrow \left[ {\matrix{ {3x – \alpha = \alpha + k2\pi } \cr {3x – \alpha = – \alpha + k2\pi } \cr } } \right. \Leftrightarrow \left[ {\matrix{ {x = {{2\alpha } \over 3} + k{{2\pi } \over 3}} \cr {x = k{{2\pi } \over 3}} \cr } } \right. \cr} \)