（2）设k(t)=f(t)-g(t) =sin(2x+π/3)+sin(2x-π/3)-√3cos2x = 2sin2xcosπ/3-√3cos2x =sin2x-√3cos2x =2(1/2sin2x-√3/2cos2x) =2sin(2x-π/3) ∴k(t)min=-2,k(t)max=2即|MN|max=2
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