Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
*)\\
y = 2 + 3\sin 4x\\
- 1 \le \sin 4x \le 1 \Leftrightarrow - 3 \le 3\sin 4x \le 3\\
\Rightarrow - 1 \le 2 + 3\sin 4x \le 5\\
\Rightarrow - 1 \le y \le 5\\
\Rightarrow \left\{ \begin{array}{l}
{y_{\min }} = - 1 \Leftrightarrow \sin 4x = - 1 \Leftrightarrow 4x = - \frac{\pi }{2} + k2\pi \Leftrightarrow x = - \frac{\pi }{8} + \frac{{k\pi }}{2}\\
{y_{\max }} = 5 \Leftrightarrow \sin 4x = 1 \Leftrightarrow 4x = \frac{\pi }{2} + k2\pi \Leftrightarrow x = \frac{\pi }{8} + \frac{{k\pi }}{2}
\end{array} \right.\\
*)\\
y = 5\cos 10x - 1\\
- 1 \le \cos 10x \le 1 \Rightarrow - 5 \le 5\cos 10x \le 5\\
\Rightarrow - 6 \le 5\cos 10x - 1 \le 4\\
\Rightarrow - 6 \le y \le 4\\
\Rightarrow \left\{ \begin{array}{l}
{y_{\min }} = - 6 \Leftrightarrow \cos 10x = - 1 \Leftrightarrow 10x = \pi + k2\pi \Leftrightarrow x = \frac{\pi }{{10}} + \frac{{k\pi }}{5}\\
{y_{\max }} = 4 \Leftrightarrow \cos 10x = 1 \Leftrightarrow 10x = k2\pi \Leftrightarrow x = \frac{{k\pi }}{5}
\end{array} \right.\\
*)\\
y = 2{\cos ^2}8x - 19\\
- 1 \le \cos 8x \le 1 \Rightarrow 0 \le {\cos ^2}8x \le 1\\
\Rightarrow - 19 \le 2{\cos ^2}8x - 19 \le - 17\\
\Rightarrow \left\{ \begin{array}{l}
{y_{\min }} = - 19 \Leftrightarrow {\cos ^2}8x = 0 \Leftrightarrow 8x = \frac{\pi }{2} + k\pi \Leftrightarrow x = \frac{\pi }{{16}} + \frac{{k\pi }}{8}\\
{y_{\max }} = - 17 \Leftrightarrow {\cos ^2}8x = 1 \Rightarrow {\sin ^2}8x = 0 \Rightarrow 8x = k\pi \Leftrightarrow x = \frac{{k\pi }}{8}
\end{array} \right.\\
*)\\
y = 6 - {\sin ^2}10x\\
- 1 \le \sin 10x \le 1 \Rightarrow 0 \le {\sin ^2}10x \le 1\\
\Rightarrow 5 \le 6 - {\sin ^2}10x \le 6\\
\Rightarrow \left\{ \begin{array}{l}
{y_{\min }} = 5 \Leftrightarrow {\sin ^2}10x = 1 \Rightarrow {\cos ^2}10x = 0 \Rightarrow 10x = \frac{\pi }{2} + k\pi \Rightarrow x = \frac{\pi }{{20}} + \frac{{k\pi }}{{10}}\\
{y_{\max }} = 6 \Leftrightarrow {\sin ^2}10x = 0 \Leftrightarrow 10x = k\pi \Leftrightarrow x = \frac{{k\pi }}{{10}}
\end{array} \right.
\end{array}\)
