解一道数学题(三角函数）

求什么？
是不是求最值？

y=cos[2(x-π/3)]+2sin(x-π/3)
=1-2[sin(x-π/3)]^2+2sin(x-π/3)
=-2[sin(x-π/3)]^2+2sin(x-π/3)+1
令a=sin(x-π/3)
则-1<=a<=1
则y=-2a^2+2a+1
=-2(a-1/2)^2+3/2
所以sin(x-π/3)=a=1/2时，y最大=3/2
sin(x-π/3)=a=-1时，y最小=-3

Y=cos(2x-2∏/3)+sin〔90-((2x-2∏/3)〕=1