Đáp án:
\(Max = 1\)
Giải thích các bước giải:
\(\begin{array}{l}
y = co{s^2}x - cosx + 1\\
Đặt:cosx = t\left( {t \in \left[ { - 1;1} \right]} \right)\\
Pt \to {t^2} - t + 1 = {t^2} - 2.t.\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{3}{4}\\
= {\left( {t - \dfrac{1}{2}} \right)^2} + \dfrac{3}{4}\\
Thay:\cos x = t\\
\to {\left( {\cos x - \dfrac{1}{2}} \right)^2} + \dfrac{3}{4}\\
Do: - 1 \le \cos x \le 1\\
\to - \dfrac{1}{2} \le \cos x - \dfrac{1}{2} \le \dfrac{1}{2}\\
\to 0 \le {\left( {\cos x - \dfrac{1}{2}} \right)^2} \le \dfrac{1}{4}\\
\to \dfrac{3}{4} \le {\left( {\cos x - \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} \le 1\\
\to Max = 1 \Leftrightarrow \cos x = 1 \to x = k2\pi \left( {k \in Z} \right)\\
Min = \dfrac{3}{4} \Leftrightarrow \cos x - \dfrac{1}{2} = 0 \Leftrightarrow \cos x = \dfrac{1}{2} \to \left[ \begin{array}{l}
x = \dfrac{\pi }{3} + k2\pi \\
x = - \dfrac{\pi }{3} + k2\pi
\end{array} \right.\left( {k \in Z} \right)
\end{array}\)
