Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
A = 3 + \dfrac{1}{4}.\sin x.\cos x = 3 + \dfrac{1}{8}.\left( {2\sin x.\cos x} \right) = 3 + \dfrac{1}{8}\sin 2x\\
- 1 \le \sin 2x \le 1 \Rightarrow - \dfrac{1}{8} \le \dfrac{1}{8}\sin 2x \le \dfrac{1}{8}\\
\Leftrightarrow - \dfrac{1}{8} + 3 \le 3 + \dfrac{1}{8}\sin 2x \le 3 + \dfrac{1}{8}\\
\Leftrightarrow \dfrac{{23}}{8} \le A \le \dfrac{{25}}{8}\\
\Rightarrow \left\{ \begin{array}{l}
{A_{\min }} = \dfrac{{23}}{8} \Leftrightarrow \sin 2x = - 1 \Leftrightarrow 2x = - \dfrac{\pi }{2} + k2\pi \Leftrightarrow x = - \dfrac{\pi }{4} + k\pi \\
{A_{\max }} = \dfrac{{25}}{8} \Leftrightarrow \sin 2x = 1 \Leftrightarrow 2x = \dfrac{\pi }{2} + k2\pi \Leftrightarrow x = \dfrac{\pi }{4} + k\pi
\end{array} \right.
\end{array}\)
