- LG a
- LG b
- LG c
- LG d
Giải các phương trình sau:
LG a
\[3{\cot ^2}\left[ {x + {\pi \over 5}} \right] = 1\]
Lời giải chi tiết:
\[\begin{array}{l}
3{\cot ^2}\left[ {x + \frac{\pi }{5}} \right] = 1\\
\Leftrightarrow {\cot ^2}\left[ {x + \frac{\pi }{5}} \right] = \frac{1}{3}\\
\Leftrightarrow \left[ \begin{array}{l}
\cot \left[ {x + \frac{\pi }{5}} \right] = \frac{1}{{\sqrt 3 }}\\
\cot \left[ {x + \frac{\pi }{5}} \right] = - \frac{1}{{\sqrt 3 }}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x + \frac{\pi }{5} = \frac{\pi }{3} + k\pi \\
x + \frac{\pi }{5} = - \frac{\pi }{3} + k\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{{2\pi }}{{15}} + k\pi \\
x = - \frac{{8\pi }}{{15}} + k\pi
\end{array} \right.
\end{array}\]
Vậy phương trình có nghiệm \[x = {{2\pi } \over {15}} + k\pi ,x = -{{8\pi } \over {15}} + k\pi \].
LG b
\[{\tan ^2}\left[ {2x - {\pi \over 4}} \right] = 3\]
Lời giải chi tiết:
\[\begin{array}{l}
{\tan ^2}\left[ {2x - \frac{\pi }{4}} \right] = 3\\
\Leftrightarrow \left[ \begin{array}{l}
\tan \left[ {2x - \frac{\pi }{4}} \right] = \sqrt 3 \\
\tan \left[ {2x - \frac{\pi }{4}} \right] = - \sqrt 3
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x - \frac{\pi }{4} = \frac{\pi }{3} + k\pi \\
2x - \frac{\pi }{4} = - \frac{\pi }{3} + k\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
2x = \frac{{7\pi }}{{12}} + k\pi \\
2x = - \frac{\pi }{{12}} + k\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{{7\pi }}{{24}} + \frac{{k\pi }}{2}\\
x = - \frac{\pi }{{24}} + \frac{{k\pi }}{2}
\end{array} \right.
\end{array}\]
Vậy \[x = {{7\pi } \over {24}} + k{\pi \over 2},x = - {\pi \over {24}} + k{\pi \over 2}\]
LG c
\[7\tan x - 4\cot x = 12\]
Lời giải chi tiết:
ĐK:
\[\begin{array}{l}
\left\{ \begin{array}{l}
\sin x \ne 0\\
\cos x \ne 0
\end{array} \right.\\
\Leftrightarrow \sin x\cos x \ne 0\\
\Leftrightarrow 2\sin x\cos x \ne 0\\
\Leftrightarrow \sin 2x \ne 0\\
\Leftrightarrow 2x \ne k\pi \\
\Leftrightarrow x \ne \frac{{k\pi }}{2}
\end{array}\]
Khi đó,
\[\begin{array}{l}
PT \Leftrightarrow 7\tan x - \frac{4}{{\tan x}} = 12\\
\Leftrightarrow 7{\tan ^2}x - 12\tan x - 4 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\tan x = 2\\
\tan x = - \frac{2}{7}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \arctan 2 + k\pi \\
x = \arctan \left[ { - \frac{2}{7}} \right] + k\pi
\end{array} \right.[TM]
\end{array}\]
Vậy \[x =\arctan 2 + k\pi ,\] \[x =\arctan \left[ { - \frac{2}{7}} \right] + k\pi\]
LG d
\[{\cot ^2}x + \left[ {\sqrt 3 - 1} \right]\cot x - \sqrt 3 = 0\]
Lời giải chi tiết:
\[\begin{array}{l}
PT \Leftrightarrow {\cot ^2}x + \sqrt 3 \cot x - \cot x - \sqrt 3 = 0\\
\Leftrightarrow \cot x\left[ {\cot x + \sqrt 3 } \right] - \left[ {\cot x + \sqrt 3 } \right] = 0\\
\Leftrightarrow \left[ {\cot x + \sqrt 3 } \right]\left[ {\cot x - 1} \right] = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cot x + \sqrt 3 = 0\\
\cot x - 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\cot x = - \sqrt 3 \\
\cot x = 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - \frac{\pi }{6} + k\pi \\
x = \frac{\pi }{4} + k\pi
\end{array} \right.
\end{array}\]
Vậy \[x = {\pi \over 4} + k\pi ,x = - {\pi \over 6} + k\pi \]