设X∈(0,3/π）,闭区间,求函数y=cos(2x-3/π)+2sin(x-6/π)的最值

X∈(0,3/π）x-π/6∈[-π/6,π/6]y=cos[2(x-π/6)]+2sin(x-π/6)=1-2sin^2(x-π/6)+2sin(x-π/6)令sin(x-π/6)=t (-1/2≤ t≤1/2)y=-2t2+2t+1=3-2(t-1/2)^2t=1/2,y最大=3t=-1/2,y最小=1