Giải phương trình sau: sinx+sin2x+sin3x=cosx+cos2x+cos3x

sinx+sin2x+sin3x=cosx+cos2x+cos3x

<=> sin3x +sinx +sin2x = cos3x +cosx +cos2x

<=> sinx(2cosx +1) = cosx(2cosx +1)

<=> (sinx - cosx)(2cosx +1)=0

<=> [ √2.sin(x-45)](2cosx +1)=0

<=> x= ± $\frac{2}{3}$ $\pi$ +k2$\pi$ ; x=$\frac{\pi}{8}$+k$\frac{\pi}{2}$