(1)
F(x)=2(cosx)^2-2√3cosxsinx=cos2x-√3sin2x+1=2sin(2x-π/6)+1
最小正周期T=π
函数f(x)在【0,π】上的单调递增区间为[0,π/3]∪[5π/6,π]
X=5π/6时,最小值f(x)min=-1
(2)
∵F(A)=2sin(2A-π/6)+1=-1
∴sin(2A-π/6)=-2,2A-π/6=3π/2,A=5π/6
sinA=1/2 cosA=-√3/2由正弦定理和余弦定理,a=2RsinA=R
(c^2+b^2-a^2)/2bc=cosA=-√3/2
(b-2c)/(acos(π/3)+C)
最后的C是大写的还是小写的?