1,函数y=cos2xsin2x+根号3*cos^22x-(根号3/2)的最小正周期为2,使m^2+2m-sinx=o（x∈R成立的实数m满足的条件为

Question

1,函数y=cos2xsin2x+根号3*cos^2 2x-(根号3/2)的最小正周期为
2,使m^2+2m-sinx=o（x∈R成立的实数m满足的条件为

Answer 1

（1）y=cos2xsin2x+√3cos^2 2x-(√3/2)
=(1/2)sin4x+√3×[(2cos^2 2x-1)+1]/2-(√3/2)
=(1/2)sin4x+(√3/2)cos4x
=sin(4x+∏/3)
∴T=2∏/4=∏/2
(2) m^2+2m-sinx=o,则m^2+2m=sinx
若要方程有根,则要y=m^2+2m与y=sinx图像有交点即可
∵当x∈R时,sinx∈[-1,1]
∴-1≤m^2+2m≤1
解-1≤m^2+2m得,m∈R
解m^2+2m≤1,得（-√2-1）≤m≤√2-1
取交集,得,m的取值范围是
[-√2-1, √2-1]