Lời giải:
1.
\(y=\sin ^4x+\cos ^4x=(\sin ^2x+\cos ^2x)^2-2\sin ^2x\cos ^2x\)
\(=1-\frac{1}{2}(2\sin x\cos x)^2=1-\frac{1}{2}\sin ^22x\)
Vì \(\sin 2x\in [-1;1]\Rightarrow \sin ^22x\in [0;1]\)
Do đó:\(y=1-\frac{1}{2}\sin ^22x\in [\frac{1}{2}; 1]\) hay \(y_{\min}=\frac{1}{2}; y_{\max}=1\)
2.
\(y=\frac{\sin x}{\cos x+2}\Rightarrow y^2=\frac{\sin ^2x}{(\cos x+2)^2}=\frac{1-\cos ^2x}{(\cos x+2)^2}\)
Đặt \(\cos x=t(t\in [-1;1])\) . Xét \(f(t)=\frac{1-t^2}{(t+2)^2}\)
\(f'(t)=\frac{-2(2t+1)}{(t+2)^3}=0\Leftrightarrow t=-\frac{1}{2}\)
Lập BBT ta suy ra \(f(t)_{\max}=f(\frac{-1}{2})=\frac{1}{3}\)
\(\Rightarrow y^2\leq \frac{1}{3}\Rightarrow \frac{-1}{\sqrt{3}}\leq y\leq \frac{1}{\sqrt{3}}\)
Vậy \(y_{\min}=\frac{-1}{\sqrt{3}}; y_{\max}=\frac{1}{\sqrt{3}}\)
