Biết \[A,\,\,B,\,\,C\] là các góc của tam giác \[ABC,\] khi đó.              

Chọn D

Đặt \(t = \tan \frac{x}{2} = \frac{1}{2}\) nên \(\sin x = \frac{{2t}}{{1 + {t^2}}} = \frac{{2\frac{1}{2}}}{{1 + \frac{1}{4}}} = \frac{4}{5}\), \(\cos x = \frac{{1 - {t^2}}}{{1 + {t^2}}} = \frac{{1 - \frac{1}{4}}}{{1 + \frac{1}{4}}} = \frac{3}{5}\).

Vậy \(\frac{{\sin x}}{{2 - 3\cos x}} = \frac{{\frac{4}{5}}}{{2 - \frac{9}{5}}} = 4\).

Cho \(\sin x = \frac{1}{5},\frac{\pi }{2} < x < \pi \). Tính \(\cot 2x\).

\(\begin{array}{l}\frac{\pi }{2} < x < \pi  \Rightarrow \frac{{2\pi }}{2} < 2x < 2\pi  \Rightarrow \pi  < 2x < 2\pi  \Rightarrow \left\{ {\begin{array}{*{20}{l}}{\cos 2x > 0}\\{\tan 2x < 0}\end{array}} \right.\\\cos 2x = 1 - 2{\sin ^2}x = 1 - 2 \cdot \frac{1}{{25}} = \frac{{23}}{{25}}\\\frac{\pi }{2} < x < \pi  \Rightarrow \cos x < 0\\\sin x = \frac{1}{5} \Rightarrow \cos x =  - \sqrt {1 - {{\sin }^2}x}  =  - \frac{{2\sqrt 6 }}{5}.\\\sin 2x = 2\sin x \cdot \cos x = 2 \cdot \frac{1}{5} \cdot \left( { - \frac{{2\sqrt 6 }}{5}} \right) =  - \frac{{4\sqrt 6 }}{5}\\\cot 2x = \frac{{\cos 2x}}{{\sin 2x}} = \frac{{\frac{{23}}{{25}}}}{{ - \frac{{4\sqrt 6 }}{5}}} =  - \frac{{23\sqrt 6 }}{{120}}\end{array}\)