已知函数f（x）=-sin²x-sinx+a（1）当f（x）=0有实数解时,求a的取值范围（2）若x∈[6/π,3/2π],

（1）令t=sinx,则f(x)=f(t)=-t^2-t+a=-(t+1/2)^2+a+1/4当f（x）=0时-(t+1/2)^2+a+1/4=0-(t+1/2)^2=-（a+1/4）(t+1/2)^2=a+1/4a+1/4≥0a≥-1/4解法2：当-t^2-t+a=0时有实数解时,依据△=b^2-4ac=1+4a≥0,得4a≥-1a≥-1...