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

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

\(\Leftrightarrow 9{{\cos }^{2}}\alpha ={{\sin }^{2}}\alpha +2\sin \alpha +1\Leftrightarrow 9\left( 1-{{\sin }^{2}}\alpha \right)={{\sin }^{2}}\alpha +2\sin \alpha +1\) \(\Leftrightarrow 10{{\sin }^{2}}\alpha +2\sin \alpha -8=0\Leftrightarrow \left[ \begin{align} & \sin \alpha =-1 \\ & \sin \alpha =\frac{4}{5} \\ \end{align} \right.\)

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

\(\bullet \) \(sin \alpha =\frac{4}{5}\Rightarrow \cos \alpha =\frac{3}{5}\xrightarrow{{}}\tan \alpha =\frac{\sin \alpha }{\cos \alpha }=\frac{4}{3}.\)