Ta có:

$\sin \left( {2\alpha + \beta } \right) = 3\sin \beta $ $ \Rightarrow \sin 2\alpha \cos \beta + \cos 2\alpha \sin \beta = 3\sin \beta $

$ \Rightarrow 2\sin \alpha \cos \alpha \cos \beta + \left( {2{{\cos }^2}\alpha - 1} \right)\sin \beta = 3\sin \beta $

$ \Rightarrow 2\sin \alpha \cos \alpha \cos \beta + 2{\cos ^2}\alpha \sin \beta = 4\sin \beta $

$ \Rightarrow 2\cos \alpha \left( {\sin \alpha \cos \beta + \sin \beta \cos \alpha } \right) = 4\sin \beta $

$ \Rightarrow \cos \alpha \sin \left( {\alpha + \beta } \right) = 2\sin \beta $

Lại có:

$\sin \left( {2\alpha + \beta } \right) = 3\sin \beta $ $ \Rightarrow \sin 2\alpha \cos \beta + \cos 2\alpha \sin \beta = 3\sin \beta $

$ \Rightarrow 2\sin \alpha \cos \alpha \cos \beta + \left( {1 - 2{{\sin }^2}\alpha } \right)\sin \beta = 3\sin \beta $

$ \Rightarrow 2\sin \alpha \cos \alpha \cos \beta - 2{\sin ^2}\alpha \sin \beta = 2\sin \beta $

$ \Rightarrow 2\sin \alpha \left( {\cos \alpha \cos \beta - \sin \beta \sin \alpha } \right) = 2\sin \beta $

$ \Rightarrow \sin \alpha \cos \left( {\alpha + \beta } \right) = \sin \beta $

Từ đó suy ra \(\dfrac{{\cos \alpha \sin \left( {\alpha + \beta } \right)}}{{\sin \alpha \cos \left( {\alpha + \beta } \right)}} = \dfrac{{2\sin \beta }}{{\sin \beta }}\) hay $\cot \alpha \tan \left( {\alpha + \beta } \right) = 2$$ \Rightarrow \tan \left( {\alpha + \beta } \right) = 2\tan \alpha $