f(x)=√3sin2x-2sin^2 x=√3sin2x+cos2x-1=2(√3/2sin2x+1/2cos2x)-1=2(sin2xcosπ/6+cos2xsinπ/6)-1=2sin(2x+π/6)-1所以当sin(2x+π/6)=1时有最大值f(x)=2-1=1f(x)=2sin(2x+π/6)-1=02sin(2x+π/6)=1sin(2x+π/6)...