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求函数y=lg(tanx+1)+根号(4sin平方x-3)的定义域
人气:228 ℃ 时间:2019-08-21 07:26:06
解答
tanx+1>0
tanx>-1
所以kπ-π/44(sinx)^2-3>=0
(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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