原式=(1+√14)/4
∵cos(x+π/6)=1/4
∴sin(x+π/6)=√15/4
∵cos(5π/6-x)=cos[π-(x+π/6)]=cos(x+π/6)
∴cos(5π/6-x)=1/4
∵cos2(π/3-x)=cos^2 (π/3-x)-sin^2 (π/3-x)
cos(π/3-x)=cos[π/2-(x+π/6)]=sin(x+π/6)=√15/4
sin(π/3-x)=sin[π/2-(x+π/6)]=cos(x+π/6)=1/4
∴cos2(π/3-x)=(√15/4)^2-(1/4)^2=√14/4
∴cos(5π/6-x)+cos2(π/3-x)=1/4+√14/4=(1+√14)/4
应该没出错,