一、 令u=2x-π/3
①当f(x)max=3时
u=π/2 + 2kx
解得:{x|x=kπ+ 5π/12}
②当f(x)min=-1时
u=3π/2 + 2kx
解得:{x|x=11π/12+kπ}
二、
若解单调性也是利用换元和还原.
令u=2x-π/3
原式等于:
f(x)=1-2sin(u)··············通过对图形的压缩和向上平移,这些对该函数的周期不影响
∴f(x)的单调递减的区间是(-π/2 +2kπ,π/2 +2kπ)···········(因为g(x)=2sinx 与f(x)=- 2sinx关于x轴对称,∴取g(x)的单调递减区间,就是f(x)的单调递增区间)
∴-π/2 +2kπ≤u≤π/2 +2kπ
解得:f(x)的单调递减区间为(-π/12 +kπ,5π/12 +kπ) 注:此括号为方括号
若-π/12 +kπ=π/4
解得:k=1/3 代入5π/12 +kπ ···············看是否大于π/2,若大于则该区间为单调减,否则反之
带入之后,5π/12 +π/3= 3π/4
3π/4>π/2
此区间为(π/4,3π/4),所以(π/4,π/2)为单调递减区间