实数x,y满足x2=2xsin(xy)-1,则x2010+6sin5y=________.
7
分析:由(x-sin(xy))2+cos2(xy)=0,知x=sin(xy),且cos(xy)=0,所以x=sin(xy)=±1.由此能求出x2010+6sin5y.
解答:(x-sin(xy))2+cos2(xy)=0
∴x=sin(xy) 且 cos(xy)=0
∴x=sin(xy)=±1
∴siny=1 xsin(xy)=1
x2010+6sin5y=7.
故答案为:7.
点评:本题考查函数的值的求法,解题时要认真审题,仔细解答,注意三角函数性质的灵活运用.