fx=sin2x/sinx+2sinx
=2sinxcosx/sinx + 2sinx
=2(cosx + sinx)
=2√2[（√2/2)cosx + (√2/2)sinx]
= 2√2sin(x + π/4)
所以函數(shù)f(x)的定義域是 x ∈R,且 x ≠kπ (k∈Z)
函數(shù)的最小正周期是 2π/1 = 2π
(2) 若 f(a) = 2,則有 2√2sin(a +π/4) = 2
即有：sin(a + π/4 ) = √2/2
因?yàn)?a∈（0,π)
a + π/4 = 3π/4
所以 a = π/2
所以 f(a+π/12) = f(7π/12) =2√2sin(7π/12 + π/4) = 2√2sin(5π/6) = 2√2×(1/2) =√2