f(x)=cos^2x+2cosx x∈[π/6,3π/4] 求值域

f(x)=cos^2x+2cosx
=（cosx+1）^2-1
x∈[π/6,3π/4]
cosx∈[—二分之根号二,二分之根号3]
cosx+1∈[1-二分之根号二,1+二分之根号3]
（cosx+1）^2∈[(1-二分之根号二）的平方,（1+二分之根号3）的平方]
再减去1就是值域了~

f（x）=cos^2x+2cosx
=（1/2）*(1+cos2x)+2cosx
=(1/2)+(1/2)*(cos^2 x - sin^2 x)+2cosx
=(1/2)+(1/2)*(2cos^2 x -1)+2cosx
=cos^2 x + 2cosx
=(cos^2 x + 2cosx + 1) - 1
=(1 + cosx)^2 - 1