Tìm min max của hs:
y=3cos4x+4sin2x3sin4x+2cos2x
Ta có y=3(1sin2x)2+4sin2x3sin4x+2(1sin2x)
Đặt t=sin2x;0t1. Khi đó y=3t22t+33t22t+2=1+13t22t+2
Xét hàm số
    f(t)=3t22t+2(0t1)
f(t)=6t2=0t=13
Ta có bảng biến thiên như hình vẽ:


Suy ra 53f(t)385y43
Vậy miny=43khi sin2x=1, chẳng hạn khi x=π2
maxy=85khi sin2x=13
Viết lại y1=3(cos4xsin4x)+4sin2x2]cos2x3sin4x+2cos2x
y1=3(cos2xsin2x+4sin2x2cos2x3sin4x+2cos2x
y1=13sin4x+2cos2x. Đặt sin2x=t,t[0;1], hàm số trở thành
  y1=13t22t+2. Gọi f(t)=3t22t+2
Thấy rằng {f(t)>0,t[0;1],doΔ=5<0vàa=3>0b2a=13[0;1], suy ra:
* minDf(t)=Δa=53maxD(y1)=35maxDy=35+1=85, có được khi và chỉ khi t=13sin2x=13
* maxDf(t)=max{f(0);f(1)}=max{2;3}=f(1)=3minD(y1)=13
minDy=13+1=43, có được khi và chỉ khi t=1sin2x=1
Tóm lại: GTLN của hàm số bằng 85; GTNN của hàm số bằng 43

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