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f(x)=2cos2x+2√3sinxcosx
=2cos2x+√3sin2x
=2(1/2cos2x+√3/2sin2x)
=2(cosπ/3cos2x+sinπ/3sin2x)
=2cos(π/3-2x)
因为-π/6<=x<=π/3
-π/3<=-x<=π/6
-2π/3<=-2x<=π/3
-π/3<=π/3-2x<=2π/3
所以,-1/2<=cos(π/3-2x)<=1/2
-1<=2cos(π/3-2x)<=1
即值域为[-1,1]
=2cos2x+√3sin2x
=2(1/2cos2x+√3/2sin2x)
=2(cosπ/3cos2x+sinπ/3sin2x)
=2cos(π/3-2x)
因为-π/6<=x<=π/3
-π/3<=-x<=π/6
-2π/3<=-2x<=π/3
-π/3<=π/3-2x<=2π/3
所以,-1/2<=cos(π/3-2x)<=1/2
-1<=2cos(π/3-2x)<=1
即值域为[-1,1]
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