高一数学，如图，第二问怎么做。
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f(x)=2cos(π/2-x)cosx-√3cos2x
=2sinxcosx-√3cos2x
=sin2x-√3cos2x
=2(1/2sin2x-√3/2cos2x)
=2sin(2x-π/3)
x在[0,π/2]
2x-π/3在[-π/3,2π/3]
sin(2x-π/3)在[-π/3,2π/3]上值域为:[-√3/2,1]
f(x)=2sin(2x-π/3)在[-π/3,2π/3]上值域为:[-√3,2]
f(x)=2sin(2x-π/3)最小值=-√3
f(x)=2sin(2x-π/3)最大值=2
=2sinxcosx-√3cos2x
=sin2x-√3cos2x
=2(1/2sin2x-√3/2cos2x)
=2sin(2x-π/3)
x在[0,π/2]
2x-π/3在[-π/3,2π/3]
sin(2x-π/3)在[-π/3,2π/3]上值域为:[-√3/2,1]
f(x)=2sin(2x-π/3)在[-π/3,2π/3]上值域为:[-√3,2]
f(x)=2sin(2x-π/3)最小值=-√3
f(x)=2sin(2x-π/3)最大值=2
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