函数题，求快速解答

1+1/2cosx≥sin^2(x/2) →
1+(1/2)(1-2sin^2 x)≥sin^2(x/2);
3/2≥2·sin^2(x/2);
则sin^2(x/2)≤3/4.
-√3/2≤sin(x/2)≤√3/2;
若满足x属于0到π,则x/2属于0到π/2.
结合上式得.x/2属于0到π/3.

f(x)
=sinx+√3cosx
=2·[(1/2)sinx+(√3/2)cosx]
=2·[cos(π/3)sinx+sin(π/3)cosx]
=2·sin(x+π/3).

x/2属于0到π/3,则x+π/3属于π/3到2π/3.
则当x+π/3=π/2,即x=π/6时,
f(x)取得最大值,为2×1=2

1+1/2*cosa≥(sinx/2)^2.===>2+cosx≥2(sinx/2)^2.===>2cosx+1≥0,===>cosx≥-1/2.===>0≤x≤2π/3.===>π/3≤x+π/3≤π.===>0≤sin(x+π/3)≤1.===>0≤2sin(x+π/3)≤2.====>0≤sinx+√3cosx≤2.===>0≤f(x)≤2.===>f(x)max=2.