函数f(x)=(1/2)cos2x-sinx+1,x∈(π/6,2π/3)的值域
问题描述:
函数f(x)=(1/2)cos2x-sinx+1,x∈(π/6,2π/3)的值域
答
f(x)=(1/2)cos2x-sinx+1
=)=(1/2)[1-2(sinx)^2]-sinx+1
=-(sinx)^2-sinx+3/2
因为x∈(π/6,2π/3)
1/2