1.
\(y=sin^4x+cos^4x=1-\dfrac{1}{2}sin^22x\)
Vì \(x\in\left[0;\dfrac{\pi}{6}\right]\Rightarrow2x\in\left[0;\dfrac{\pi}{3}\right]\Rightarrow t=sin2x\in\left[\dfrac{1}{2};\dfrac{\sqrt{3}}{2}\right]\)
Khi đó: \(y=f\left(t\right)=1-\dfrac{1}{2}t^2\)
\(\Rightarrow minf\left(t\right)=min\left\{f\left(\dfrac{1}{2}\right);f\left(\dfrac{\sqrt{3}}{2}\right);f\left(0\right)\right\}=f\left(\dfrac{\sqrt{3}}{2}\right)=\dfrac{5}{8}\)
\(maxf\left(t\right)=max\left\{f\left(\dfrac{1}{2}\right);f\left(\dfrac{\sqrt{3}}{2}\right);f\left(0\right)\right\}=f\left(0\right)=1\)
2.
\(y=2sin^2x-cos2x=4sin^2x-1\)
\(x\in\left[0;\dfrac{\pi}{3}\right]\Rightarrow t=sinx\in\left[0;\dfrac{\sqrt{3}}{2}\right]\)
Khi đó: \(y=f\left(t\right)=4t^2-1\)
\(\Rightarrow minf\left(t\right)=min\left\{f\left(0\right);f\left(\dfrac{\sqrt{3}}{2}\right)\right\}=f\left(0\right)=-1\)
\(maxf\left(t\right)=max\left\{f\left(0\right);f\left(\dfrac{\sqrt{3}}{2}\right)\right\}=f\left(\dfrac{\sqrt{3}}{2}\right)=2\)
