Question

已知函数f(x)=2√3sin²x-sin(2x-π/3)

(1)求函数f(x)的最小正周期及单调增区间；
(2)设α∈（0，π），f(α/2)=1/2+√3,求sinα的值；
(3)若x∈[-π/2,0],函数f(x)的最大值

Answer 1

(1)f(x)=√3(1-cos2x)-1/2sin2x+√3/2cos2x
=√3-1/2sin2x-√3/2cos2x
=√3-sin(2x+π/3)
∴最小正周期T=2π/2=π
单调增区间：π/2+2kπ<=2x+π/3<=3π/2+2kπ
即π/12+kπx<=x<=7π/12+kπ
(2)f(α/2)=1/2+√3=√3-sin(α+π/3)
∴sin(α+π/3)= -1/2
α+π/3= -π/6+2kπ或-5π/6+2kπ
α= -π/2+2kπ或 -7π/6+2kπ
∵α∈（0，π）∴α=5π/6
∴sinα=1/2
(3)x∈[-π/2,-5π/12]时，f(x)单调增；
x∈[-5π/12,0]时，单调减。
∴f(x)max=f(-5π/12)=√3+1

Answer 2

弱弱的问一句是2倍根号3乘sinx的平方么
f(x)=√3(1-cos2x)-1/2sin2x+√3/2cos2x
=√3-1/2sin2x-√3/2cos2x
=√3-sin(2x+π/3)
∴最小正周期T=2π/2=π
单调增区间：π/2+2kπ<=2x+π/3<=3π/2+2kπ
即π/12+kπx<=x<=7π/12+kπ
(2)f(α/2)=1/2+√3=√3-sin(α+π/3)
∴sin(α+π/3)= -1/2
α+π/3= -π/6+2kπ或-5π/6+2kπ
α= -π/2+2kπ或 -7π/6+2kπ
∵α∈（0，π）∴α=5π/6
∴sinα=1/2
(3)x∈[-π/2,-5π/12]时，f(x)单调增；
x∈[-5π/12,0]时，单调减。
∴f(x)max=f(-5π/12)=√3+1