Question

这个三角函数题目求解答，过程手写，谢谢

Answer 1

(1).f(x)=(1/2)(1+cos2x) - (√3/2)sin2x+1
=(1/2)cos2x-(√3/2)sin2x+(3/2)
=cos(2x+π/3)+(3/2)
T=(2π/2)= π
由0+2kπ≤(2x+π/3) ≤π+2kπ得：
-π/6+kπ≤x≤π/3+kπ
单调增区间为：[-π/6+kπ , π/3+kπ]
(2)
cos(2θ+π/3)+(3/2)=(5/6)
cos(2θ+π/3)= - (2/3)
π≤2θ+π/3≤5π/3
sin(2θ+π/3)= - (√5/3)
sin2θ=sin[(2θ+π/3)-π/3]=sin(2θ+π/3)cos(π/3)-cos(2θ+π/32θ+π/3)sin(π/3)
=(-√5/3)(1/2)-(-2/3)(√3/2)
=(2√3-√5)/6