f(x)=√3（1-cosx)+sinx-√3=2sin(x-π/3) 2kπ+π/2≤x-π/3≤2kπ+3π/2 k∈z 解得
2kπ+5π/6≤x≤ 2kπ+11π/6 k∈z所以单调递减区间是[ 2kπ+5π/6,2kπ+11π/6] k∈z
(2) x∈[-π/2,π/2]时,-5π/6≤x-π/3≤π/6 ,所以当x=-π/6时f(x)有最小值是-2；当x=π/2是f(x)有最大值是2