2 câu trả lời
$$\eqalign{ & {\cos ^2}x + {1 \over {{{\cos }^2}x}} = 2\left( {\cos x - {1 \over {\cos x}}} \right) + 1 \cr & DK:\,\,\cos x \ne 0 \Leftrightarrow x \ne {\pi \over 2} + k\pi \cr & Dat\,\,t = \cos x - {1 \over {\cos x}} \Rightarrow {t^2} = {\cos ^2}x + {1 \over {{{\cos }^2}x}} - 2 \cr & \Rightarrow {\cos ^2}x + {1 \over {{{\cos }^2}x}} = {t^2} + 2 \cr & PT:\,\,{t^2} + 2 = 2t + 1 \Leftrightarrow {t^2} - 2t + 1 = 0 \Leftrightarrow t = 1 \cr & \Rightarrow \cos x - {1 \over {\cos x}} = 1 \Leftrightarrow {\cos ^2}x - \cos x - 1 = 0 \cr & \Leftrightarrow \left[ \matrix{ \cos x = {{1 + \sqrt 5 } \over 2}\,\,\left( {loai} \right) \hfill \cr \cos x = {{1 - \sqrt 5 } \over 2} \hfill \cr} \right. \Leftrightarrow x = \pm \arccos {{1 - \sqrt 5 } \over 2} + k2\pi \,\,\left( {k \in Z} \right) \cr} $$
