函数f(x)=-sin平方x+sinx+a,若1<=f(x)<=17/4,对一切x属于R恒成立,求a的取值范围

F(X)=-SIN^2X+SINX+A
=-(sinx-1/2)^2+A+1/4
因为:-1<=sinx<=1,所以-3/2<=sinx-1/2<=1/2,
-9/4<=-(sinx-1/2)^2<=0,
a-2<=f(x)<=a+1/4

又有：1=<F(X)=<17/4
则:
1<=a-2,
a+1/4<=17/4
解得
a>=3,
a<=4
实数A取值范围3<=A<=4.

为什么:-1<=sinx<=1?