f(x)=2sin(π/6-2x)
=1/2
则:sin(π/6-2x)=1/4
则:sin(5π/6+2x)+sin^2(π/3+2x)
=sin[π-(π/6-2x)]+{sin[π/2-(π/6-2x)]}^2
=sin(π/6-2x)+[cos(π/6-2x)]^2
=sin(π/6-2x)+[1-sin^2(π/6-2x)]
=1/4+1-(1/4)^2
=19/16
若f(x)=2sin(π/6-2x),当f(x)=1/2时,求sin(5π/6+2x)+sin²(π/3+2x
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