已知f(x)=(sinx)^2+2sinxcosx+3(cosx)^2,x∈R,求:
(1)函数f(x)的最大值及取得最大值的自变量x的集合
(2)f(x)的单调增区间
参考答案:这才是普通的解法
(1)f(x)=1+2sinxcosx+2(cosx)^2
=1+sin2x+cos2x+1
=根号2sin(2x+π/4)+2
最大值 根号2+2
2x+π/4=2kπ+π/2 x=kπ+π/8
最大值 -根号2+2
2x+π/4=2kπ+3π/2 x=kπ+5π/8
(2)2kπ-π/2<=2x+π/4<=2kπ+π/2
kπ-3π/8<=x<=kπ+π/8