tanx>-1
所以kπ-π/4
(sinx)^2>=3/4
-1<=sinx<=-√3/2,√3/2<=sinx<=1
所以2kπ+4π/3<=x<=2kπ+5π/3
2kπ+π/3<=x<=2kπ+2π/3
即kπ+π/3<=x<=kπ+2π/3
所以是kπ+π/3<=x
求函数y=lg（tanx+1）+根号（4sin平方x-3）的定义域
求函数y=lg（tanx+1）+根号（4sin平方x-3）的定义域
数学人气：440 ℃时间：2019-08-21 07:26:06
优质解答
tanx+1>0
tanx>-1
所以kπ-π/4
(sinx)^2>=3/4
-1<=sinx<=-√3/2,√3/2<=sinx<=1
所以2kπ+4π/3<=x<=2kπ+5π/3
2kπ+π/3<=x<=2kπ+2π/3
即kπ+π/3<=x<=kπ+2π/3
所以是kπ+π/3<=x
tanx>-1
所以kπ-π/4
(sinx)^2>=3/4
-1<=sinx<=-√3/2,√3/2<=sinx<=1
所以2kπ+4π/3<=x<=2kπ+5π/3
2kπ+π/3<=x<=2kπ+2π/3
即kπ+π/3<=x<=kπ+2π/3
所以是kπ+π/3<=x
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