Question

已知函数f(x)=2√3sinx·cosx+cos^2x-sin^2x-1(x∈R)

(1)求单调递增区间
(2)若x∈[-5pai/12,pai/3]
第二步要求详细点

Answer 1

答：
f(x)=2√3sinxcosx+cos²x-sin²x-1
=√3sin2x+cos2x-1
=2*[sinx(√3/2)+cos2x(1/2)-1
=2sin(x+π/6)-1

（1）f(x)=2sin(x+π/6)-1的单调增区间满足：-π/2+2kπ<=x+π/6<=π/2+2kπ
所以：函数的单调增区间为[2kπ-2π/3，2kπ+π/3]，k∈Z

（2）若-5π/12<=x<=π/3
则：-π/4<=x+π/6<=π/2
所以：-√2/2<=sin(x+π/6)<=1
所以：-√2<=2sin(x+π/6)<=2
所以：-1-√2<=f(x)<=1
所以：f(x)的值域为[-1-√2，1]