Question

y=根号3 sin2x - 2sin²x 求函数的最大值 求函数的零点集合

Answer 1

y=√3sin2x-2(1-cos2x)/2
=2(√3/2*sin2x+cos2x*1/2)-1
=2(sin2xcosπ/6+cos2xsinπ/6)-1
=2sin(2x+π/6)-1
所以最大值=2-1=1

y=0
则sin(2x+π/6)=1/2
2x+π/6=2kπ+π/6，2x+π/6=2kπ+5π/6
x=kπ，x=kπ+π/3
所以零点集合{x|x=kπ，x=kπ+π/3，k∈Z}

Answer 2

y=√3sin2x-2sin^2x
=√3sin2x+cos2x-1
=2(√3/2sin2x+1/2cos2x)-1
=2sin(2x+π/6)-1
所以当sin(2x+π/6)=1时有最大值2-1=1
当y=0时
则sin(2x+π/6)=1/2
2x+π/6=2kπ+π/6 或 2x+π/6=2kπ+5π/6
x=kπ或 x=kπ+π/3

Answer 3

y=√3sin2x-2sin²x
=√3sin2x+cos2x-1
=2(√3/2sin2x+1/2cos2x)-1
=2sin(2x+π/6)-1

根据图形 最大值是1 函数的零点集合为x属于kπ K为整数

Answer 4

x=k/2π-π/12