Cho biết \(3\cos \alpha  - \sin \alpha  = 1\), \({0^0} < \alpha  < {90^0}.\) Giá trị của \(\tan \alpha \) bằng   

Ta có \(3\cos \alpha  - \sin \alpha  = 1 \Leftrightarrow 3\cos \alpha  = \sin \alpha  + 1\)\( \Rightarrow 9{\cos ^2}\alpha  = {\left( {\sin \alpha  + 1} \right)^2}\)

\( \Leftrightarrow 9{\cos ^2}\alpha  = {\sin ^2}\alpha  + 2\sin \alpha  + 1\) \( \Leftrightarrow 9\left( {1 - {{\sin }^2}\alpha } \right) = {\sin ^2}\alpha  + 2\sin \alpha  + 1\)

\( \Leftrightarrow 10{\sin ^2}\alpha  + 2\sin \alpha  - 8 = 0 \Leftrightarrow \left[ \begin{array}{l}\sin \alpha  =  - 1\\\sin \alpha  = \dfrac{4}{5}\end{array} \right..\)

\( \bullet \) \(\sin \alpha  =  - 1\): không thỏa mãn vì \({0^0} < \alpha  < {90^0}.\)

\( \bullet \) \(\sin \alpha  = \dfrac{4}{5} \Rightarrow \cos \alpha  = \dfrac{3}{5}\)\( \Rightarrow \tan \alpha  = \dfrac{{\sin \alpha }}{{\cos \alpha }} = \dfrac{4}{3}\)