f(x)=sinx-2(cos(x/2))^2=sinx-(1+cosx)=sinx-cosx-1=√2*sin(x-π/4)-1令2kπ-π/2＜x-π/4＜2kπ+π/2(k∈Z)得2kπ-π/4＜x＜2kπ+3π/4(k∈Z)所以f(x)的一个单调递增区间是(2kπ-π/4,2kπ+3π/4)(k∈Z)同理也可...