Question

已知函数fx=cos(2x+3分之派)+2sinx的平方,求f(x)在x属于【负派,0】上递减区间

Answer 1

f(x)=cos(2x+π/3)+2sin^2(x)
=cos(2x)cos(π/3)-sin(2x)sin(π/3)+[1-cos(2x)]
=(1/2)cos(2x)-(√3/2)sin2x+1-cos(2x)
= - [(√3/2)sin2x+(1/2)cos(2x)]+1
= - sin(2x+π/6)+1
在正弦函数前面加了一负号后单调性与没有负号的单调性相反,
欲求单调减区间须代入到单调增区间中去解出x ;
由 -π/2+2kπ≤2x+π/6≤π/2+2kπ得：
-2π/3+2kπ≤2x≤π/3+2kπ
-π/3+kπ≤x≤π/6+kπ
当k= - 1时,
-4π/3≤x≤ -5π/6,再与[-π,0]求交集得：[-π, -5π/6]
当k=0时,
-π/3≤x≤π/3与[-π , 0 ]求交集得：[-π/3 , 0]
所以原函数的单调减区间为：
[-π, -5π/6] ; [-π/3 , 0]