f(x)=5√3*cos²x+√3*sin²x+4sinx*cosx-3√3
=4√3*cos²x+4sinx*cosx-2√3
=2√3*(2cos²x-1)+2*2sinx*cosx
=2√3*cos2x+2*sin2x
=4*(√3/2*cos2x+1/2*sin2x)
=4(cos2x*sinπ/3+sin2x*sinπ/3)
=4*sin(2x+π/3).（你化错了）
最小正周期：T=2π/2=π,
x∈[0,π]时,
2x+π/3∈[π/3,7π/3],
f(x)=4*sin(2x+π/3)∈[-4,4].
f(x)最大值：4,此时x=π/12；
最小值：-4,此时x=7π/12.
因为y=sinx (x∈R),的单调递增区间为：[2kπ-π/2,2kπ+π/2],
由2kπ-π/2