... ()()()() 11 1 1 1 cos 6x 1 cos 8x 1 cos10x 1 cos12x22 2 2−−+=− −+ ⇔ cos 6x cos 8x cos10x cos12x+= +⇔ 2cos7xcosx 2cos11xcosx=⇔ ( )2cos x cos7x cos11x 0−= ⇔ cos x 0 cos7x cos11x=∨ ... ≠⎨⎪≠⎩Do đó : (*)⇔ 222 11 1 11 1cos x sin x sin 2x 3⎛⎞⎛⎞⎛ ⎞−+ −+ −=⎜⎟⎜⎟⎜ ⎟⎝⎠⎝⎠⎝ ⎠ 11 ⇔22 22 11 1 cos x sin x 4 sin x cos x 3++ =20 ⇔22224sin x 4cos x 1 204sin xcos x 3++= ... ∨=+π∨=π+π ∈2kxxkx2,55 2kZ Bài 3 1: Giải phương trình ( )22 2 2sin x sin 3x cos 2x cos 4x *+=+ Ta có (*) ⇔ ()()()() 11 11 1 cos 2x 1 cos 6x 1 cos4x 1 cos 8x2222−+−=+++ ⇔ ()cos2x...