求y=2cos^x+2sinx-4在[-2π/3,2π/3] 的值域

x∈[-2π/3,2π/3]
sinx∈【-1,1】
令sinx=t,t ∈【-1,1】
y=2(1-t²)+2t-4
=-2t²+2t-2
=-2(t-1/2)²-3/2
t=1/2,最大值为-3/2
t=-1,最小值为-6
值域【-6,-3/2】

y=2cos^x+2sinx-4=-2sin^x+2sinx-2=-2(sinx-1/2)^2-3/2
在区间[-2π/3,2π/3]sinx可取到[-1，1]
y==-2(sinx-1/2)^2-3/2<=-3/2
sinx=-1时
y=-2(sinx-1/2)^2-3/2>=-2（-1-1/2)^2-3/2=-6