Học lớp hướng dẫn giải
Ta có \(f'\left( x \right) = 2\cos 2x - 2cox = 0 \Leftrightarrow \left[ \begin{array}{l} cos = 1\\ \cos = - \frac{1}{2} \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = k2\pi \\ x = \pm \frac{{2\pi }}{3} + k2\pi \end{array} \right.\left( {k \in Z} \right)\)
\(\Rightarrow \left\{ \begin{array}{l} f\left( {k2\pi } \right) = 0\\ f\left( {\frac{{2\pi }}{3} + k2\pi } \right) = - \frac{{3\sqrt 3 }}{2}\\ f\left( { - \frac{{2\pi }}{3} + k2\pi } \right) = \frac{{3\sqrt 3 }}{2} \end{array} \right. \Rightarrow Max{\rm{ }}f\left( x \right) = f\left( { - \frac{{2\pi }}{3} + k2\pi } \right) = \frac{{3\sqrt 3 }}{2}.\)
