f(x)=2sin(x/4)cos(x/4)-2根号3*sin²(x/4)+根号3 求1.当x∈[0,π],求f(x)值域2.f(x)单调递增区间
问题描述:
f(x)=2sin(x/4)cos(x/4)-2根号3*sin²(x/4)+根号3 求1.当x∈[0,π],求f(x)值域2.f(x)单调递增区间
答
f(x)=2sin(x/4)cos(x/4)-2v3*sin²(x/4)+v3 =sin(x/2)-v3*(1-cos(x/2))+v3 =sin(x/2)+v3cos(x/2) =2sin(x/2 +π/3)x∈[0,π]. ...