+) ta có nếu : \(cos\alpha=0\) thì \(pt\Leftrightarrow sin^2x=-\dfrac{1}{8}\left(vôlí\right)\) (vì \(sin^2\alpha+cos^2\alpha=1\)) \(\Rightarrow cosx\ne0\)
+) ta có : \(cos^2\alpha-2sin^2\alpha=\dfrac{1}{4}\) \(\Leftrightarrow1-2tan^2\alpha=\dfrac{1}{4}\left(\dfrac{1}{cos^2\alpha}\right)\)
\(\Leftrightarrow1-2tan^2\alpha=\dfrac{1}{4}\left(1+tan^2\alpha\right)\) \(\Leftrightarrow\dfrac{9}{4}tan^2\alpha-\dfrac{3}{4}\Leftrightarrow tan^2\alpha=\dfrac{1}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}tan\alpha=\dfrac{1}{\sqrt{3}}\\tan\alpha=\dfrac{-1}{\sqrt{3}}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}tan\alpha=tan30\\tan\alpha=tan\left(-30\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\alpha=30+k360\\\alpha=-30+k360\end{matrix}\right.\)
