Đáp án:
\(B = - 36 - 16\sqrt 5 \)
Giải thích các bước giải:
\(\begin{array}{l}
Do:x \in \left( {\pi ;\dfrac{{3\pi }}{2}} \right)\\
\to \left\{ \begin{array}{l}
\cos x < 0\\
\sin x < 0
\end{array} \right.\\
\tan x - \cot x = \dfrac{3}{2}\\
\to \tan x - \dfrac{1}{{\tan x}} = \dfrac{3}{2}\\
\to \dfrac{{2{{\tan }^2}x - 3\tan x - 2}}{{2\tan x}} = 0\left( {\tan x \ne 0} \right)\\
\to 2{\tan ^2}x - 3\tan x - 2 = 0\\
\to \left[ \begin{array}{l}
\tan x = 2\\
\tan x = - \dfrac{1}{2}
\end{array} \right.\\
\to \left[ \begin{array}{l}
\sin x = 2\cos x\\
\sin x = - \dfrac{1}{2}\cos x \to \cos x = - 2\sin x
\end{array} \right.\\
Do:{\sin ^2}x + {\cos ^2}x = 1\\
\to \left[ \begin{array}{l}
4{\cos ^2}x + {\cos ^2}x = 1\\
4{\sin ^2}x + {\sin ^2}x = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
{\cos ^2}x = \dfrac{1}{5}\\
{\sin ^2}x = \dfrac{1}{5}
\end{array} \right. \to \left[ \begin{array}{l}
\cos x = - \dfrac{1}{{\sqrt 5 }} \to \sin x = - \dfrac{2}{{\sqrt 5 }}\\
\sin x = - \dfrac{1}{{\sqrt 5 }} \to \cos x = \dfrac{1}{{2\sqrt 5 }}\left( {KTM} \right)
\end{array} \right.\\
\to B = \dfrac{{2.2\cos x - 2}}{{\cos x + \dfrac{1}{2}}} = \left( {4\cos x - 2} \right):\left( {\dfrac{{2\cos x + 1}}{2}} \right)\\
\to B = \dfrac{{8\cos x - 4}}{{2\cos x + 1}}\\
Thay:\cos x = - \dfrac{1}{{\sqrt 5 }} \to B = - 36 - 16\sqrt 5
\end{array}\)
