Question

已知函数f(x)=3cos^2x+2cosx*sinx+sin^2x 求详细解答过程

（1）求函数的周期
（2）写出f(x)的单调区间
（3）求f(x)的最大值，病求出此时x的值

Answer 1

f(x)=3(cosx)^3+2sinxcosx+(sinx)^2
=sin2x+2(cosx)^2+1
=sin2x+cos2x+2
=√2sin(2x+π/4)+2

(1)最小正周期为T=2π/2=π，周期为kπ，k是不为0的整数。

(2)2kπ-π/2<2x+π/4<2kπ+π/2，则kπ-3π/8<x<kπ+π/8，单调递增区间为(kπ-3π/8,kπ+π/8)。
2kπ+π/2<2x+π/4<2kπ+3π/2，则kπ+π/8<x<kπ+5π/8，单调递减区间为(kπ+π/8,kπ+5π/8)。

(3)当2x+π/4=2kπ+π/2，即x=kπ+π/8时，f(x)取得最大值为√2+2。

Answer 2

解：
f(x)=3cos^2x+2cosx*sinx+sin^2x
=2cos^2x+sin2x+1
=cos2x+sin2x+2
=√2*sin(2x+π/4)+2
∴T=2π/2=π
2kπ-π/2≤2x+π/4≤2kπ+π/2时，f(x)单调递增 x∈[kπ-3π/8，kπ+π/8]
∴单调增区间为[kπ-3π/8，kπ+π/8]
单调减区间为[kπ+π/8，kπ+5π/8]

2x+π/4=2kπ+π/2
x=kπ+π/8时
f(x)max=2+√2

Answer 3

同上。。。