y=sin(-1/2x+π/3)的单调递增区间得到单调增为2kπ-3π/2≤-1/2x+π/3≤2kπ+π/2,k∈Z得出x的值,当x∈（0,π],时,2kπ-3π/2>0,2kπ+π/2≤π,设f(x)=-1/2x+π/3,为单调函数,且最大值为f(π）,可得出k值,从而得出x,...