Question

用三角函数线解sin(cosx)<=0

Answer 1

设t=cosx
因-1<=cosx<=1，即-1<=t<=1
所以当-1<=t<=0时，t为第四象限角，所以sint=sin(cosx)<=0
即-1<=cosx<=0时，sin(cosx)<=0
此时根据余弦线，知
2kπ+π/2<=x<=2kπ+3π/2 ,k∈Z。

Answer 2

解：
设t=cosx
因-1<=cosx<=1，即-1<=t<=1
所以当-1<=t<=0时，t为第四象限角，所以sint=sin(cosx)<=0
即-1<=cosx<=0时，sin(cosx)<=0
此时根据余弦线，知
2kπ+π/2<=x<=2kπ+3π/2 ,k∈Z。

Answer 3

设cosx=m,-1<=m<=1，解sinm<=0,得-1<=m<=0，所以-1<=cosx<=0，
所以，2kπ+1/2π<=x<=2kπ+3/2π,记得写成集合形式啊。