The general solution of sin x - 3sin 2x + sin 3x = cos x -3cos 2x + cos 3x is

∵sin⁡x−3sin⁡2x+sin⁡3x=cos⁡x−3cos⁡2x+cos⁡3x(sin⁡3x+sin⁡x)−3sin⁡2x=(cos⁡3x+cos⁡x)−3cos⁡2x2sin⁡2x⋅cos⁡x−3sin⁡2x=2cos⁡2x⋅cos⁡x−3cos⁡2xsin⁡2x(2cos⁡x−3)−cos⁡2x(2cos⁡x−3)=0(sin⁡2x−cos⁡2x)(2cos⁡x−3)=0So, sin⁡2x−cos⁡2x=0as  2cos⁡x−3≠0now, sin⁡2x=cos⁡2xtan⁡2x=1=tan⁡π42x=nπ+π4x=nπ2+π8