Đáp án: GTNN y=5/2 và GTLN y=11/2
Giải thích các bước giải:
$\begin{array}{l}
y = {\sin ^6}x + co{s^6}x + 3\\
= {\left( {{{\sin }^2}x} \right)^3} + {\left( {co{s^2}x} \right)^3} + 3\\
= \left( {{{\sin }^2}x + co{s^2}x} \right)\left( {{{\sin }^4}x - {\mathop{\rm s}\nolimits} {\rm{inx}}.c{\rm{osx + co}}{{\rm{s}}^2}x} \right) + 3\\
= 1.\left( {{{\left( {{{\sin }^2}x + c{\rm{o}}{{\rm{s}}^2}x} \right)}^2} - 3{\mathop{\rm s}\nolimits} {\rm{inx}}.c{\rm{osx}}} \right) + 3\\
= 1 - \frac{3}{2}.2{\mathop{\rm s}\nolimits} {\rm{inx}}.c{\rm{osx}} + 3\\
= 4 - \frac{3}{2}.\sin 2x\\
Do: - 1 \le \sin 2x \le 1\\
\Rightarrow \frac{5}{2} \le 4 - \frac{3}{2}.\sin 2x \le \frac{{11}}{2}\\
\Rightarrow \frac{5}{2} \le y \le \frac{{11}}{2}\\
\Rightarrow \left\{ \begin{array}{l}
GTNN:y = \frac{5}{2} \Leftrightarrow \sin 2x = 1 \Rightarrow x = \frac{\pi }{4} + k\pi \\
GTLN:y = \frac{{11}}{2} \Leftrightarrow \sin 2x = - 1 \Rightarrow x = - \frac{\pi }{4} + k\pi
\end{array} \right.
\end{array}$
