解不等式tan(2x+π/3)>(根号3)/3

问题描述:

解不等式tan(2x+π/3)>(根号3)/3

tan(π/6)=√3/3
tan周期是π
所以tan(kπ+π/6)=√3/3
tan在一个周期(kπ-π/2,kπ+π/2)内是增函数
tan(2x+π/3)>tan(kπ+π/6)
所以kπ+π/6kπ+π/6-π/3kπ-π/6所以kπ/2-π/12