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<2x+π/6<π/2
=> cos(2x+π/6)=2√2/3
sin(2x)=sin(2x+π/6-π/6)
=sin(2x+π/6)cos(π/6)-cos(2x+π/6)sin(π/6)
=(1/3)*√3/2-(2√2/3)*1/2
=(√3-2√2)/6

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