令x'=2x+π/6, 则f(x')=2sinx'。由正弦函数的增减性知:
f(x')在[0,π/2]内为增函数,在[π/2,3π/2]内为减函数。
即当0<=x'<=π/2时f(x')是增函数,当π/2<=x'<=3π/2时f(x')是减函数
由0<=2x+π/6<=π/2得,0<=x<=π/6, 即f(x)在[0,π/6]内为增函数
由π/2<=2x+π/6<=3π/2得,π/6<=x<=2π/3,即f(x)在[π/6,2π/3]内为减函数。
而[π/6,π,2]是[π/6,2π/3]的一个子区间,故f(x)在[π/6,π/2]内为减函数.
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