2010浙江高考数学 选择题 求解过程，网上答案不够详细
推荐回答（5个）
把f(x)看成是p(x)=4sin（2x+1）和q(x)=-x的迭加.划出两个简图,你会发现当x=-4时fx)>0,x=-2时,f(x)>0,在-4
同理可分析其他情况.由于用作图法较简单,所以是数形结合
画y=x与y=4sin（2x+1） 图像。。看交点是哪个区间
这样的，f(x)=0时，函数f（x）=4sin（2x+1）-x=0，函数g（x）=4sin（2x+1）和u(x)=-x
分别画两个函数的图像。g（x）=4sin（2x+1）的图像有sinx图像经过平移转换得到的。（这叫转化思想）两个图像的交点就是0点。看看落在哪个区间就行了。（画图和计算结合就叫数形结合。）
f(x)的零点就是h(x)=4sin(2x+1)和g(x)=x两个函数图象的交点，在同一坐标系中作出两个函数图象，很容易发现h(x)=4sin（2x+1）在（-4，-2）大于0，而g(x)=x在（-4，-2）小于0，两函数图象没有交点，所以函数f（x）=4sin（2x+1）-x在（-4，-2）不存在零点
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