f(x)=2cosxsin(x+π/3）-√3sin^2x+sinxcos
=sin(x+π/3+x）+sin(x+π/3-x）-√3(1-cos^2x)+sinxcos
=sin(x+π/3+x）+sin(π/3）-√3+√3cos^2x+0.5sin2x
=sin(2x+π/3）+√3/2-√3+√3cos^2x+0.5sin2x
=sin(2x+π/3）-√3/2+√3*(1+cos2x)/2+0.5sin2x
=sin2x*cos(π/3）+cos2x*sin(π/3）+(√3/2)cos2x+0.5sin2x
=0.5sin2x+(√3/2)cos2x+(√3/2)cos2x+0.5sin2x
=sin2x+(√3)cos2x
=(2/2)*[sin2x+(√3)cos2x]
=2*[(1/2)*sin2x+(√3/2)cos2x]
=2*[sin(π/6)*sin2x+cos(π/6)cos2x]
=2cos(2x-π/6)
1.求函數(shù)f(x)得值域
1≥cos(2x-π/6)≥-1
2≥f(x)≥-2
到這里你應該會了