f(x)=2cos2x+2sqrt(3)sin2x=4sin(2x+Pi/6)周期T=2Pi/2=Pi,解方程2x+Pi/6=Pi/2,得x=Pi/6解方程2x+Pi/6=-Pi/2,得x=-Pi/3所以,f(x)在[kPi-Pi/3,kPi+Pi/6]单调递增,[kPi+Pi/6,kPi+2Pi/3]单调递减,(k属于整数集)在范围[-Pi/3,Pi/4]时，当x=Pi/6,有最大值4
