设函数,f(x)=2cos^2x+√3sin2x,若f(x)=5/3,-π/6<x<π/6,求sin2x
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/06 19:35:05
![设函数,f(x)=2cos^2x+√3sin2x,若f(x)=5/3,-π/6<x<π/6,求sin2x](/uploads/image/z/7020484-52-4.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0%2Cf%EF%BC%88x%EF%BC%89%3D2cos%5E2x%2B%E2%88%9A3sin2x%2C%E8%8B%A5f%EF%BC%88x%EF%BC%89%3D5%2F3%2C-%CF%80%2F6%EF%BC%9Cx%EF%BC%9C%CF%80%2F6%2C%E6%B1%82sin2x)
设函数,f(x)=2cos^2x+√3sin2x,若f(x)=5/3,-π/6<x<π/6,求sin2x
设函数,f(x)=2cos^2x+√3sin2x,若f(x)=5/3,-π/6<x<π/6,求sin2x
设函数,f(x)=2cos^2x+√3sin2x,若f(x)=5/3,-π/6<x<π/6,求sin2x
f(x)=2cos^2x+√3sin2x=cos2x+1+√3sin2x
=2sin(2x+π/6)+1=5/3
=> sin(2x+π/6)=1/3
-π/2
cos(2x)=cos^2(x)-sin^2(x)=2cos^2(x)-1
<===> 2cos^2(x) = cos(2x) + 1
f(x)=2cos^2(x)+ Root(3)sin(2x)
=cos(2x) + 1+ Root(3)sin(2x)
=2{(1/2)*cos(2x) + (Root(3)/2)*sin(2x)} + 1
...
全部展开
cos(2x)=cos^2(x)-sin^2(x)=2cos^2(x)-1
<===> 2cos^2(x) = cos(2x) + 1
f(x)=2cos^2(x)+ Root(3)sin(2x)
=cos(2x) + 1+ Root(3)sin(2x)
=2{(1/2)*cos(2x) + (Root(3)/2)*sin(2x)} + 1
=2sin(2x + 30) + 1
f(x)=5/3
<===> 2sin(2x+30)=5/3 - 1
<===> 2sin(2x+30)=2/3
<===> sin(2x+30)=1/3
cos(2x+30) = Root( 1-sin^2(2x+30) )
= Root( 8/9 ) >0
Becasuse of -30 < x < 30 ===> cos(x) > 0
sin(2x)=sin(2x+30-30)
=sin(2x+30)cos(30)-cos(2x+30)sin(30)
=(1/3)*(Root(3)/2) - Root(8/9)*(1/2)
=Root(3)/6 - Root(8)/6
收起