Giải thích các bước giải: Ta có: \(\begin{array}{l}y = {\left( {1 – {{\cos }^2}x} \right)^2} – 2{\cos ^2}x + 1\\ = 1 – 2{\cos ^2}x + {\cos ^4}x – 2{\cos ^2}x + 1\\ = {\cos ^4}x – 4{\cos ^2}x + 2\\ = \left( {{{\cos }^4}x – 4{{\cos }^2}x + 4} \right) – 2\\ = {\left( {{{\cos }^2}x – 2} \right)^2} – 2\\ – 1 \le \cos x \le 1 \Rightarrow 0 \le {\cos ^2}x \le 1\\ \Rightarrow – 2 \le {\cos ^2}x – 2 \le – 1\\ \Leftrightarrow 1 \le {\left( {{{\cos }^2}x – 2} \right)^2} \le 4\\ \Leftrightarrow – 1 \le {\left( {{{\cos }^2}x – 2} \right)^2} – 2 \le 2\\ \Leftrightarrow – 1 \le y \le 2\\ \Rightarrow \left\{ \begin{array}{l}{y_{\min }} = – 1 \Leftrightarrow {\cos ^2}x = 1 \Rightarrow {\sin ^2}x = 0 \Leftrightarrow x = k\pi \\{y_{\max }} = 2 \Leftrightarrow {\cos ^2}x = 0 \Leftrightarrow x = \frac{\pi }{2} + k\pi \end{array} \right.\end{array}\) Reply
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
y = {\left( {1 – {{\cos }^2}x} \right)^2} – 2{\cos ^2}x + 1\\
= 1 – 2{\cos ^2}x + {\cos ^4}x – 2{\cos ^2}x + 1\\
= {\cos ^4}x – 4{\cos ^2}x + 2\\
= \left( {{{\cos }^4}x – 4{{\cos }^2}x + 4} \right) – 2\\
= {\left( {{{\cos }^2}x – 2} \right)^2} – 2\\
– 1 \le \cos x \le 1 \Rightarrow 0 \le {\cos ^2}x \le 1\\
\Rightarrow – 2 \le {\cos ^2}x – 2 \le – 1\\
\Leftrightarrow 1 \le {\left( {{{\cos }^2}x – 2} \right)^2} \le 4\\
\Leftrightarrow – 1 \le {\left( {{{\cos }^2}x – 2} \right)^2} – 2 \le 2\\
\Leftrightarrow – 1 \le y \le 2\\
\Rightarrow \left\{ \begin{array}{l}
{y_{\min }} = – 1 \Leftrightarrow {\cos ^2}x = 1 \Rightarrow {\sin ^2}x = 0 \Leftrightarrow x = k\pi \\
{y_{\max }} = 2 \Leftrightarrow {\cos ^2}x = 0 \Leftrightarrow x = \frac{\pi }{2} + k\pi
\end{array} \right.
\end{array}\)