LG a

\(2{\tan ^2}x + 3 = {3 \over {\cos x}}\)

Lời giải chi tiết:

\(\begin{array}{l}
PT \Leftrightarrow 2.\frac{{{{\sin }^2}x}}{{{{\cos }^2}x}} + 3 = \frac{3}{{{{\cos }^2}x}}\\
\Leftrightarrow 2.\frac{{1 - {{\cos }^2}x}}{{{{\cos }^2}x}} + 3 = \frac{3}{{{{\cos }^2}x}}\\
\Leftrightarrow 2\left( {\frac{1}{{{{\cos }^2}x}} - 1} \right) + 3 = \frac{3}{{{{\cos }^2}x}}
\end{array}\)

Đặt \(t = {1 \over {\cos x}}\left( {x \ne {\pi  \over 2} + k\pi } \right)\) Ta có: \(\eqalign{  & 2\left( {{t^2} - 1} \right) + 3 = 3t \cr &\Leftrightarrow 2{t^2} - 3t + 1 = 0  \cr  &  \Leftrightarrow \left[ {\matrix{   {t = 1}  \cr   {t = {1 \over 2}}  \cr  } } \right. \Leftrightarrow \left[ {\matrix{   {\cos x = 1}  \cr   {\cos x = 2\,\left( \text{loại} \right)}  \cr  } } \right. \cr &\Leftrightarrow x = k2\pi  \cr} \)

Cách khác:

LG b

\({\tan ^2}x = {{1 + \cos x} \over {1 + \sin x}}\)

Lời giải chi tiết:

Điều kiện : \(\cos x \ne 0 \Leftrightarrow x = {\pi  \over 2} + k\pi \) \(\eqalign{  & {\tan ^2}x = {{1 + \cos x} \over {1 + \sin x}} \cr &\Leftrightarrow {{{{\sin }^2}x} \over {{{\cos }^2}x}} = {{1 + \cos x} \over {1 + \sin x}}  \cr  &  \Leftrightarrow {{1 - {{\cos }^2}x} \over {1 - {{\sin }^2}x}} = {{1 + \cos x} \over {1 + \sin x}}  \cr  &  \Leftrightarrow \frac{{1 - {{\cos }^2}x}}{{\left( {1 - \sin x} \right)\left( {1 + \sin x} \right)}} = \frac{{1 + \cos x}}{{1 + \sin x}}\cr & \Leftrightarrow {{1 - {{\cos }^2}x} \over {1 - \sin x}} = 1 + \cos x \cr &(Do\, 1+\sin x\ne 0)\cr &  \Rightarrow 1 - {\cos ^2}x = \left( {1 - \sin x} \right)\left( {1 + \cos x} \right) \cr &\Leftrightarrow \left( {1 + \cos x} \right)\left( {1 - \cos x} \right) - \left( {1 - \sin x} \right)\left( {1 + \cos x} \right) = 0 \cr &\Leftrightarrow \left( {1 + \cos x} \right)\left( {1 - \cos x - 1 + \sin x} \right) = 0 \cr &\Leftrightarrow \left( {1 + \cos x} \right)\left( {\sin x - \cos x} \right) = 0 \cr &\Leftrightarrow \left[ \begin{array}{l}1 + \cos x = 0\\\sin x = \cos x\end{array} \right.\cr & \Leftrightarrow \left[ {\matrix{   {\cos x =  - 1}  \cr   {\tan x = 1}  \cr  } } \right. \cr &\Leftrightarrow \left[ {\matrix{   {x =  \pi  + k2\pi }  \cr   {x = {\pi  \over 4} + k\pi }  \cr  }\left( {k \in\mathbb Z} \right) } \right. \cr} \)

LG c

\(\tan x + \tan 2x = {{\sin 3x} \over {\cos x}}\) 

Lời giải chi tiết:

Điều kiện 

\(\left\{ \begin{array}{l}
\cos x \ne 0\\
\cos 2x \ne 0
\end{array} \right. \) \(\Leftrightarrow \left\{ \begin{array}{l}
x \ne \frac{\pi }{2} + k\pi \\
x \ne \frac{\pi }{4} + \frac{{k\pi }}{2}
\end{array} \right.\)

\(\eqalign{  & {\mathop{\rm tanx}\nolimits}  + tan2x = {{\sin 3x} \over {\cos x}} \cr & \Leftrightarrow \frac{{\sin x}}{{\cos x}} + \frac{{\sin 2x}}{{\cos 2x}} = \frac{{\sin 3x}}{{\cos 3x}} \cr & \Leftrightarrow \frac{{\sin x\cos 2x + \cos x\sin 2x}}{{\cos x\cos 2x}} = \frac{{\sin 3x}}{{\cos 3x}}\cr &\Leftrightarrow {{\sin 3x} \over {\cos x\cos 2x}} = {{\sin 3x} \over {\cos x}}  \cr  &  \Leftrightarrow \frac{{\sin 3x - \sin 3x\cos 2x}}{{\cos x\cos 2x}} = 0\cr &\Rightarrow \sin 3x - \sin 3x\cos 2x=0 \cr &\Leftrightarrow \sin 3x\left( {1 - \cos 2x} \right) = 0  \cr  &  \Leftrightarrow \left[ {\matrix{   {\sin 3x = 0}  \cr   {\cos 2x = 1}  \cr  } } \right. \Leftrightarrow \left[ {\matrix{   {\sin 3x = 0}  \cr   {\sin x = 0}  \cr  } } \right.\cr &\Leftrightarrow \left[ \begin{array}{l}x = \frac{{k\pi }}{3}\\x = k\pi \end{array} \right.\cr &\Leftrightarrow x = k{\pi  \over 3},k \in\mathbb  Z \cr} \)

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