Cho (y = sin 2x – 2cos x). Giải phương trình (y’ = 0).

Ta có (y’ = 2cos 2x + 2sin x).

(begin{array}{l}y’ = 0 Leftrightarrow 2cos 2x + 2sin x = 0 Leftrightarrow cos 2x + sin x = 0\ Leftrightarrow cos 2x =  – sin x = cos left( {x + dfrac{pi }{2}} right)\ Leftrightarrow left[ begin{array}{l}2x = x + dfrac{pi }{2} + k2pi \2x =  – x – dfrac{pi }{2} + k2pi end{array} right. Leftrightarrow left[ begin{array}{l}x = dfrac{pi }{2} + k2pi \x = dfrac{{ – pi }}{6} + dfrac{{k2pi }}{3}end{array} right.,,left( {k in mathbb{Z}} right)end{array})

Chọn B