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