$1)$ `sin (3x+(\pi)/(4))-cos (x+(2\pi)/(3))=0`

`=> cos [(\pi)/(2)-(3x+(\pi)/(4))]=cos (x+(2\pi)/(3))`

`=> cos ((\pi)/(4)-3x)=cos (x+(2\pi)/(3))`

`=> (\pi)/(4)-3x=x+(2\pi)/(3)+k2\pi` hoặc `(\pi)/(4)-3x=-x-(2\pi)/(3)+k2\pi (k in ZZ)`

`=> x=-(5\pi)/(48)-(k\pi)/(2)` hoặc `x=(11\pi)/(24)-k\pi`

Vậy `x in {-(5\pi)/(48)-(k\pi)/(2); (11\pi)/(24)-k\pi | k in ZZ}`

$2)$ `cos^2 3x+sin 3x+1=0`

`=> 1-sin^2 3x+sin 3x+1=0`

`=> sin^2 3x-sin 3x-2=0`

`=> (sin 3x-2)(sin 3x+1)=0`

`=> sin 3x=2` (loại vì `sin 3x<=1`) hoặc `sin 3x=-1 (TM)`

`=> 3x=-(\pi)/(2)+k2\pi (k in ZZ)`

`=> x=-(\pi)/(6)+(k2\pi)/(3)`

$3)$ `sin 3x+\sqrt{3}cos 3x=2sin 2x`

`=> (1)/(2)sin 3x+(\sqrt{3})/(2)cos 3x=sin 2x`

`=> cos  (\pi)/(3)sin 3x+sin  (\pi)/(3)cos 3x=sin 2x`

`=> sin (3x-(\pi)/(3))=sin 2x`

`=> 3x-(\pi)/(3)=2x+k2\pi` hoặc `3x-(\pi)/(3)=\pi-2x+k2\pi (k in ZZ)`

`=> x=(\pi)/(3)+k2\pi` hoặc `x=(4\pi)/(15)+(k2\pi)/(5)`

Vậy `x in {(\pi)/(3)+k2\pi; (4\pi)/(15)+(k2\pi)/(5) | k in ZZ}`