Đáp án:
\[{y_{\max }} = 3 \Leftrightarrow x = 2\pi + k4\pi \,\,\,\,\left( {k \in Z} \right)\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
- 1 \le \sin \left( {\dfrac{x}{2} - \dfrac{{3\pi }}{2}} \right) \le 1\\
\Leftrightarrow - 2 \le 2\sin \left( {\dfrac{x}{2} - \dfrac{{3\pi }}{2}} \right) \le 2\\
\Leftrightarrow - 2 \le - 2\sin \left( {\dfrac{x}{2} - \dfrac{{3\pi }}{2}} \right) \le 2\\
\Leftrightarrow 1 + \left( { - 2} \right) \le 1 + \left( { - 2\sin \left( {\dfrac{x}{2} - \dfrac{{3\pi }}{2}} \right)} \right) \le 1 + 2\\
\Leftrightarrow - 1 \le 1 - 2\sin \left( {\dfrac{x}{2} - \dfrac{{3\pi }}{2}} \right) \le 3\\
\Leftrightarrow - 1 \le y \le 3\\
\Rightarrow {y_{\max }} = 3 \Leftrightarrow \sin \left( {\dfrac{x}{2} - \dfrac{{3\pi }}{2}} \right) = - 1\\
\Leftrightarrow \dfrac{x}{2} - \dfrac{{3\pi }}{2} = - \dfrac{\pi }{2} + k2\pi \\
\Leftrightarrow \dfrac{x}{2} = \pi + k2\pi \\
\Leftrightarrow x = 2\pi + k4\pi \,\,\,\,\,\left( {k \in Z} \right)\\
Vậy\,\,\,{y_{\max }} = 3 \Leftrightarrow x = 2\pi + k4\pi \,\,\,\,\left( {k \in Z} \right)
\end{array}\)