Question

大神来看看下面这道数学题谢谢

Answer 1

解，f(x)=1/2-cos2x/2+√3/2sin2x
=1/2+sin2xcosπ/6-sinπ/6cos2x
=1/2+sin(2x-π/6)
则T=2π/2=π
(2)x∈(-π/3，π/6)，则2x-π/6∈(-5π/6，π/6)
则f(x)∈(-1/2，1)

Answer 2

f(x) = (sinx)^2 + √3sinxcosx = 1/2 - (1/2)cos2x + (√3/2)sin2x = 1/2 + sin(2x-π/6)
最小正周期 是 π，f(-π/3) = 0, f(-π/6) = -1/2, f(π/6) = 1, 取值范围是 -1/2≤f≤1.