Bài giải của hocmai.toanhoc( Trịnh Hào Quang)
Bài 2:
[TEX]VT = c{\rm{os}}3x\left( {\frac{{c{\rm{os}}3x + 3\cos x}}{4}} \right) - \sin 3x\left( {\frac{{3\sin x - \sin 3x}}{4}} \right) \\ = \frac{1}{4}(c{\rm{os}}^2 3x + \sin ^2 3x) + \frac{3}{4}(c{\rm{os}}xc{\rm{os}}3x - \sin x\sin 3x) \\= \frac{1}{4} + \frac{3}{4}c{\rm{os}}4x[/TEX].
Vậy:
[TEX]\frac{1}{4} + \frac{3}{4}c{\rm{os}}4x = \frac{{2 + 3\sqrt 2 }}{8} \Leftrightarrow c{\rm{os}}4x = \frac{{\sqrt 2 }}{2} \Leftrightarrow x = \pm \frac{\pi }{{16}} + k\frac{\pi }{2}(k \in Z)[/TEX]