f(x)=2√3cos�0�5x-2sinxcosx-√3=√3(2cos�0�5x-1)-2sinxcosx=√3cos2x-sin2x=2(√3/2cos2x-1/2sin2x)=2sin(π/3-2x)=-2sin(2x-π/3)(1)令π/2+2kπ≤2x-π/3≤3π/2+2kπ得5π/12+kπ≤x≤11π/12+kπ即[5π/12+kπ，11π/12+kπ]，k属于Z
f【x】=√3【cos2x+1】-sin2x-√3 =2cos【2x+π/6]-√3 2kπ-π ≤ 2x+π/6 ≤ 2kπ k属于z解得【kπ-7π/12，kπ-π/12] 不懂可以再问
[k*pi+5*pi/12,k*pi+11*pi/12][k属于Z] pi即派
[5派/12+ 2k派, 11派/12+2k派]
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