f(x)=2倍根号3*SinxCosx+2Cos²x是这样吧
=√3sin2x+cos2x+1
=2sin(2x+π/6)+1
(1)T=2π/2=π
(2)2kπ-π/2≤2x+π/6≤2kπ+π/2
2kπ-2π/3≤2x≤2kπ+π/3
kπ-π/3≤x≤kπ+π/6
增区间【kπ-π/3,kπ+π/6】,k∈Z
(3)当2x+π/6=2kπ+π/2,即x=kπ+π/6,k∈Z时,y有最大值3
和我写的差不多,但是最后一步我不会--,求解。有个2SIN要在-1和1乘以2吗然后再加1
正弦值属于【-1,1】所以,是正弦值等于1时,y有最大值。所以最大值为3.