+) ta có nếu : \(cos\alpha=0\) thì \(pt\Leftrightarrow sin^2x=-\dfrac{1}{8}\left(vôlí\right)\) (vì \(sin^2\alpha+cos^2\alpha=1\)) \(\Rightarrow cosxe0\)
+) ta có : \(cos^2\alpha-2sin^2\alpha=\dfrac{1}{4}\) \(\Leftrightarrow1-2tan^2\alpha=\dfrac{1}{4}\left(\dfrac{1}{cos^2\alpha}\right)\)