Y=log0.2[1-2sin(2x+π/3)],求定义域,值域.周期,
问题描述：
Y=log0.2[1-2sin(2x+π/3)],求定义域,值域.周期,
还有当X【0.π】,求函数最小X的值
答
定义域是1-2sin(2x+π/3)>0
sin(2x+π/3)