(1)若sin2xsin3x=cos2xcos3x,则x的一个值为( )

1.
sin2xsin3x=cos2xcos3x
--> cos2xcos3x-sin2xsin3x=0
--> cos(2x+3x)=0
--> cos5x=0
故5x=kπ+π/2(k∈Z)
x=kπ/5+π/10(k∈Z)
k=0时x=π/10,故x的一个值为π/10.
2.
设t=cosAcosB.
1/2+t=sinAsinB+cosAcosB=cos(A-B)
t-1/2=cosAcosB-sinAsinB=cos(A+B)
故-1≤1/2+t≤1,-1≤t-1/2≤1
解得-1/2≤t≤1/2,
即cosAcosB的取值范围是[-1/2,1/2].