设函数f(x)=√3cos²x+sinxcosx-√3/2（1）求函数f(x)的最小正周期T,并求出函数f(x)的单调递增区间；（2）求在[0,3π）内使f(x)取到最大值的所有x的和

Question

设函数f(x)=√3cos²x+sinxcosx-√3/2（1）求函数f(x)的最小正周期T,并求出函数f(x)的单调递增区间；
（2）求在[0,3π）内使f(x)取到最大值的所有x的和

Answer 1

(1)
f(x)=√3cos²x+sinxcosx-√3/2
=(√3/2)[cos(2x)+1]+(1/2)sin(2x)-√3/2
=(√3/2)cos(2x)+(1/2)sin(2x)
=cos(π/6)cos(2x)+sin(π/6)sin(2x)
=cos(2x-π/6)
最小正周期T=2π/2=π
2kπ-π≤2x-π/6≤2kπ时,f(x)单调递增,此时kπ-5π/12≤x≤kπ+π/12
2kπ≤2x-π/6≤2kπ+π时,f(x)单调递减,此时kπ+5π/12≤x≤kπ+7π/12
函数的单调递增区间为[kπ-5π/12,kπ+π/12],单调递减区间为[kπ+5π/12,kπ+7π/12],(k∈Z)
(2)
x∈[0,3π]
-π/6≤2x-π/6≤35π/6
当2x-π/6=0,2π,4π时,f(x)取得最大值.
对应的x分别为π/12,13π/12,25π/12,
π/12+13π/12+25π/12=(1+13+25)π/12=39π/12