已知f(x)=2sin^2(x-π/12)+（根号3)*sin(2x-π/6)，求f(x)的最大值及取得最大值时对应的x值！
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參考答案:令x-π/12 = y
则2sin^2(x-π/12)+（根号3)*sin(2x-π/6)
= 2sin^2 y+(√3)*sin(2y)
= 1-cos(2y)+(√3)*sin(2y)
= 1/2[sin(-2y +π/3)] + 1
= 1/2[sin(-2x +π/2)] + 1
再令w = sin(-2x +π/2) ∈[-1，1]
2sin^2(x-π/12)+（根号3)*sin(2x-π/6)∈[1/2,3/2]
当f(x) = 1/2时 -2x +π/2 = 2kπ-π/2 x = (π/2)-kπ (k∈z)
当f(x) = 3/2时 -2x +π/2 = 2kπ+π/2 x = -kπ (k∈z)
