Question

高中数学 函数题 求解 需要过程 谢谢

Answer 1

f(x)=sin2xcos(π/6) - cos2xsin(π/6) - 2(1 - cos2x)
=sin2x•(√3/2) - cos2x•(1/2) - 2 + 2cos2x
=(√3/2)sin2x + (3/2)cos2x - 2
=√[(√3/2)² + (3/2)²]•sin(2x+φ) - 2
其中tanφ=(3/2)/(√3/2)=√3
=√3sin(2x + π/3) - 2
(1)T=2π/2=π
(2)2kπ - π/2≤2x + π/3≤2kπ + π/2
2kπ - 5π/6≤2x≤2kπ + π/6
kπ - 5π/12≤x≤kπ + π/12，k∈Z

Answer 2

f(x)=sin(2x-π/6)-4(sinx)^2
=√3/2sin2x-1/2cos2x-2(1-cos2x)
=√3/2sin2x+3/2cos2x-2
=√3(1/2sin2x+√3/2cos2x)-2
=√3sin(2x+π/3)-2
T=2π/2=π
2x+π/3在[2kπ-π/2,2kπ-π/2]是单调递增
x在[kπ-5π/12,kπ-π/12]是单调递增