已知f(sinx)=2sin^2x-sinx,求f(x)
f(sinx)=2sin²x-sinx令sinx=t
则：
f(t)=2t²-t
即：
f(x)=2x²-x
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绛旓細f(x)=2sinxcosx+sin(2x+娲/2)=sin2x+cos2x =(鏍瑰彿2)sin(2x+(pi/4))鏈灏忔鍛ㄦ湡=2pi/2=pi 褰-(pi/2)+2n*pi<=2x+(pi/4)<=(pi/2)+2n*pi 鍗筹細n*pi -(3pi/8) <=x<= n*pi +(pi/8) 鏃讹紝f(x)鍗曡皟閫掑
