已知函数f(x)=2cos2（平方）x+2根号3sinxcosx+a且f(π/6)=4

f(x)=2(cosx)^2+2√3sinxcosx+a
=√3sin2x+cos2x+1+a
=2sin(2x+π/6)+1+a
(1)
f(π/6)=4
f(π/6)=2sin(2*π/6+π/6)+1+a=2+1+a=4
所以a=1
(2)
f(x)=2sin(2x+π/6)+2
-π/4≤x≤π/3
-π/3≤2x+π/6≤5π/6
所以-√3≤2sin(2x+π/6)≤2
所以2-√3≤f(x)≤4
故值域是[2-√3,4]
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