Question

已知函数f(x)=2cos^2x+2根号3sinxcosx

Answer 1

首先a^b表示a的b次方
y=4cosx^2
4根号3sinxcosx-2
（首先把自变量的形式都化为相同，等下好合并）
=2(cos2x
1)
2根号3sin2x-2
=2[cos2x
根号3sin2x]
=4[1/2cos2x
2分之根号3sin2x]
=4sin(π/6
2x)
所以最小正周期t=2π/2=π
这种题目注意一种形式的化简
asinx
bsinx
=根号a^2
b^2[a/(根号a^2
根号b^2)
b/(根号a^2
根号b^2)]
然后直接解出
最小正周期t=2π/w

Answer 2

答：
f(x)=2(cosx)^2+2√3sinxcosx
=cos2x+1+√3sin2x
=2*[(√3/2)sin2x+(1/2)cos2x]+1
=2sin(2x+π/6)+1
单调递增区间满足：2kπ-π/2<=2x+π/6<=2kπ+π/2
kπ-π/3<=x<=kπ+π/6
所以：单调递增区间为[kπ-π/3，kπ+π/6]，k∈Z
单调递减区间满足：2kπ+π/2<=2x+π/6<=2kπ+3π/2
kπ+π/6<=x<=kπ+2π/3
所以：单调递增区间为[kπ+π/6，kπ+2π/3]，k∈Z