Bài 1.26 trang 11 SBT Đại số và Giải tích 11 Nâng cao

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 \(x = {\pi  \over 4} + k\pi ,x =  - {\pi  \over 6} + k\pi \)

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