答:由sinx=sin2x=2sinxcosx,则cosx=1/2,所以x=π/3,
在0~π/3上,sin2x>sinx,π/3~π上,sinx>sin2x,
S=∫(0,π/3)(sin2x-sinx)dx+∫(π/3,π)(sinx-sin2x)dx
答:由sinx=sin2x=2sinxcosx,则cosx=1/2,所以x=π/3,
在0~π/3上,sin2x>sinx,π/3~π上,sinx>sin2x,
S=∫(0,π/3)(sin2x-sinx)dx+∫(π/3,π)(sinx-sin2x)dx