y=（1+cos2x,1)*(1,√3sin2x+a)=cos2x+√3sin2x+a+1=2(1/2cos2x+√3/2sin2x)+a+1=2(sinπ/6cos2x+consπ/3sin2x)+a+1=2sin(2x+π/6)+a+1y=f(x)=o在[0,3π/4]时有两个不同的实根,则f(0)≤0f(3π/4)≤0f(π/6)＞0解得...