2、由(1)如何推出(2)
2、由(1)如何推出(2)
(1)[kπ,kπ+π/3]∪[kπ+π/2,kπ+(5π)/6],k∈z
(2)[(kπ)/2,(kπ)/2+π/3],k∈z
3.由(1)如何推出(2)
(1){kπ}∪{kπ+π/2},k∈z
(2){(kπ)/2},k∈z
2、由(1)如何推出(2)
(1)[kπ,kπ+π/3]∪[kπ+π/2,kπ+(5π)/6],k∈z
(2)[(kπ)/2,(kπ)/2+π/3],k∈z
[kπ,kπ+π/3]∪[kπ+π/2,kπ+(5π)/6]=[2k(π/2),2k(π/2)+π/3]∪[2k(π/2)+π/2,2k(π/2)+π/2+π/3]]=[2k(π/2),2k(π/2)+π/3]∪[(2k+1)(π/2),(2k+1)(π/2)+π/3]=[(kπ)/2,(kπ)/2+π/3],k∈z
上式中前一部分为π/2的偶数倍，后一部分为π/2的奇数倍。
3.由(1)如何推出(2)
(1){kπ}∪{kπ+π/2},k∈z
(2){(kπ)/2},k∈z
[kπ]∪[kπ+π/2]=[2k(π/2)]∪[[(2k+1)(π/2)]=[(kπ)/2],k∈z
答：-1≤1-x^2≤1 -2≤-x^2≤0 0≤x^2≤2 ｜x｜≤√2 -√2≤x≤√2详情>>
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