cos(x+π/6)=1/3,(0

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cos(x+π/6)=1/3,(0
cos(x+π/6)=1/3,(0

cos(x+π/6)=1/3,(0
0则π/6所以sin(x+π/6)=√[1-cos²(x+π/6)]=√(1-1/9)=2√2/3
故cosx=cos[(x+π/6)-π/6]
=cos(x+π/6)cos(π/6)+sin(x+π/6)sin(π/6)
=(1/3)*(√3/2)+(2√2/3)*(1/2)
=(√3+2√2)/6