求值遇y=sin^2x+cosx+2 (pi/3

y=1-(cosx)^2+cosx+2
令a=cosx
则y=-a^2+a+3
π/3

y=sin^2x+cosx+2=-cos^2x+cosx+3
令t=cosx，t值域为[-1,1/2]
y=-t^2+t+3，定义域为t的值域[-1,1/2]
由此算得y的值域

y=sin^2x+cosx+2
=3-cos^2x+cosx
=-(cosx-1/2)^2+13/4
因为pi/3<=x<=pi，(cosx-1/2)^2>=0
故当cosx＝-1时，y有最小值＝-9/4+13/4＝1
故当cosx-1/2＝0时，y有最大值＝13/4
所以y=sin^2x+cosx+2 的值域为[1,13/4]