已知tanx=a,求使sinx+sin2x+sin3x=0时a的值

sinx+sin2x+sin3x=sin(2x-x)+sin2x+sin(2x+x)
=sin2xcosx-cos2xsinx+sin2x+sin2xcosx+cos2xsinx
=2sin2xcosx+sin2x
=(2cosx+1)sin2x
=0
所以cosx=-1/2或sin2x=0,
若sin2x=0,则2sinxcosx=0,
即sinx=0(因为tanx=a存在,所以cosx≠0),
cosx=1,tanx=0；
若cosX=-1/2,则sinX=±√3/2,
所以tanX=±√3
综上,a=0或±√3.

Sin 3x=sin(2x+x) = cos(2x)sin(x) + sin(2x)cos(x)
= (cos^2(x) - sin^2(x))sin(x) + 2cos(x)sin(x)cos(x)
= sin(x) (cos^2(x)-sin^2(x) + 2 cos^2(x))
=sin(x) (3cos^2(x) - sin^2(x))=sin(x) (4cos^2...