f=[根号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]
当2x-π/6=π/2时,f(x)取得最大值0
当2x-π/6=-π/3时,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问的∵-π/6<2C-π/6<11π/6是怎么来的0