a. Đặt \(t = {1 \over {\cos x}}\left( {x \ne {\pi \over 2} + k\pi } \right)\)
Ta có:
\(\eqalign{ & 2\left( {{t^2} - 1} \right) + 3 = 3t \Leftrightarrow 2{t^2} - 3t + 1 = 0 \cr & \Leftrightarrow \left[ {\matrix{ {t = 1} \cr {t = {1 \over 2}} \cr } } \right. \Leftrightarrow \left[ {\matrix{ {\cos x = 1} \cr {\cos x = 2\,\left( \text{loại} \right)} \cr } } \right. \Leftrightarrow x = \pi + k2\pi \cr} \)
b. Điều kiện : \(\cos x \ne 0 \Leftrightarrow x = {\pi \over 2} + k\pi \)
\(\eqalign{ & {\tan ^2}x = {{1 + \cos x} \over {1 + \sin x}} \Leftrightarrow {{{{\sin }^2}x} \over {{{\cos }^2}x}} = {{1 + \cos x} \over {1 + \sin x}} \cr & \Leftrightarrow {{1 - {{\cos }^2}x} \over {1 - {{\sin }^2}x}} = {{1 + \cos x} \over {1 + \sin x}} \cr & \Leftrightarrow {{1 - {{\cos }^2}x} \over {1 - \sin x}} = 1 + \cos x \Leftrightarrow \left[ {\matrix{ {\cos x = - 1} \cr {1 - \cos x = 1 - {\mathop{\rm sinx}\nolimits} } \cr } } \right. \cr & \Leftrightarrow \left[ {\matrix{ {\cos x = - 1} \cr {\tan x = 1} \cr } } \right. \Leftrightarrow \left[ {\matrix{ {x = \pi + k2\pi } \cr {x = {\pi \over 4} + k\pi } \cr }\left( {k \in\mathbb Z} \right) } \right. \cr} \)
c. Điều kiện \(\cos x \ne 0,\cos 2x \ne 0 \Leftrightarrow \left\{ {\matrix{ {\cos x \ne 0} \cr {{\mathop{\rm cosx}\nolimits} \ne \pm {1 \over {\sqrt 2 }}} \cr } } \right.\)
\(\eqalign{ & {\mathop{\rm tanx}\nolimits} + tan2x = {{\sin 3x} \over {\cos x}} \Leftrightarrow {{\sin 3x} \over {\cos x\cos 2x}} = {{\sin 3x} \over {\cos x}} \cr & \Leftrightarrow \sin 3x = sin3xcos2x \Leftrightarrow sin3x\left( {1 - \cos 2x} \right) = 0 \cr & \Leftrightarrow \left[ {\matrix{ {\sin 3x = 0} \cr {\cos 2x = 1} \cr } } \right. \Leftrightarrow \left[ {\matrix{ {\sin 3x = 0} \cr {\sin x = 0} \cr } } \right. \Leftrightarrow x = k{\pi \over 3},k \in\mathbb Z \cr} \)
