（1）
f=[根號(hào)3]/2 sin2x-cos^2(x)-1/2
=√3/2*sin2x-1/2(1+cos2x)-1/2
=√3/2sin2x-1/2cos2x-1
=sin(2x-π/6)-1
∵x∈[-π/12,5π/12]
∴2x∈[-π/6,5π/6]
2x-π/6∈[-π/3,2π/3]
當(dāng)2x-π/6=π/2時(shí),f(x)取得最大值0
當(dāng)2x-π/6=-π/3時(shí),f(x)取得最小值-√3/2-1
(2)
∵f(C)=sin(2C-π/6)-1=0
∴sin(2C-π/6)=1
∵-π/6第2問(wèn)的∵-π/6<2C-π/6<11π/6是怎么來(lái)的0