Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
A = \frac{{2\sin x + \cos x}}{{\cos x - 3\sin x}} = \dfrac{{\frac{{2\sin x}}{{\cos x}} + 1}}{{1 - \frac{{3\sin x}}{{\cos x}}}} = \frac{{2\tan x + 1}}{{1 - 3\tan x}} = \frac{{2.\left( { - 2} \right) + 1}}{{1 - 3.\left( { - 2} \right)}} = \frac{{ - 3}}{7}\\
b,\\
B = \frac{{2\sin x + 3\cos x}}{{3\sin x - 2\cos x}} = \dfrac{{2\frac{{\sin x}}{{\cos x}} + 3}}{{3.\frac{{\sin x}}{{\cos x}} - 2}} = \frac{{2\tan x + 3}}{{3\tan x - 2}} = \frac{{2.\left( { - 2} \right) + 3}}{{3.\left( { - 2} \right) - 2}} = \frac{1}{8}\\
c,\\
{\cos ^2}x = 1 - {\sin ^2}x = 1 - {\left( {\frac{1}{3}} \right)^2} = \frac{8}{9}\\
C = \frac{{\tan x + \cot x}}{{\tan x - \cot x}} = \dfrac{{\frac{{\sin x}}{{\cos x}} + \frac{{\cos x}}{{\sin x}}}}{{\frac{{\sin x}}{{\cos x}} - \frac{{\cos x}}{{\sin x}}}} = \dfrac{{\frac{{{{\sin }^2}x + {{\cos }^2}x}}{{\sin x.\cos x}}}}{{\frac{{{{\sin }^2}x - {{\cos }^2}x}}{{\sin x.\cos x}}}} = \frac{{{{\sin }^2}x + {{\cos }^2}x}}{{{{\sin }^2}x - {{\cos }^2}x}} = \dfrac{1}{{\frac{1}{9} - \frac{8}{9}}} =  - \frac{9}{7}\\
d,\\
\frac{1}{{{{\sin }^2}x}} = \frac{{{{\sin }^2}x + {{\cos }^2}x}}{{{{\sin }^2}x}} = 1 + \frac{{{{\cos }^2}x}}{{{{\sin }^2}x}} = 1 + {\cot ^2}x = 1 + {\left( { - 3} \right)^2} = 10\\
D = \frac{{{{\sin }^2}x + 3\sin x.\cos x - 2{{\cos }^2}x}}{{1 + 4{{\sin }^2}x}}\\
 = \dfrac{{1 + \frac{{3\cos x}}{{\sin x}} - 2.\frac{{{{\cos }^2}x}}{{{{\sin }^2}x}}}}{{\frac{1}{{{{\sin }^2}x}} + 4}}\\
 = \dfrac{{1 + 3\cot x - 2{{\cot }^2}x}}{{\frac{1}{{{{\sin }^2}x}} + 4}}\\
 = \frac{{1 + 3.\left( { - 3} \right) - 2.{{\left( { - 3} \right)}^2}}}{{10 + 4}} =  - \frac{{13}}{7}\\
e,\\
\tan x = \frac{1}{2} \Leftrightarrow \frac{{\sin x}}{{\cos x}} = \frac{1}{2} \Leftrightarrow \cos x = 2\sin x\\
E = \frac{{3{{\sin }^3}x - 2\sin x + \cos x}}{{{{\cos }^3}x + 2\sin x.{{\cos }^2}x}}\\
 = \frac{{3.{{\sin }^3}x - 2\sin x + 2\sin x}}{{{{\left( {2\sin x} \right)}^3} + 2.\sin x.{{\left( {2\sin x} \right)}^2}}}\\
 = \frac{{3{{\sin }^3}x}}{{16{{\sin }^3}x}} = \frac{3}{{16}}\\
f,\\
180^\circ  < x < 270^\circ  \Rightarrow \left\{ \begin{array}{l}
\sin x < 0\\
\cos x < 0
\end{array} \right.\\
\sin x < 0 \Rightarrow \sin x =  - \sqrt {1 - {{\cos }^2}x}  =  - \frac{3}{5}\\
F = \frac{{1 + \tan x}}{{1 - \tan x}} = \dfrac{{1 + \frac{{\sin x}}{{\cos x}}}}{{1 - \frac{{\sin x}}{{\cos x}}}} = \dfrac{{\frac{{\cos x + \sin x}}{{\cos x}}}}{{\frac{{\cos x - \sin x}}{{\cos x}}}} = \frac{{\cos x + \sin x}}{{\cos x - \sin x}} = \frac{{\frac{{ - 4}}{5} + \frac{{ - 3}}{5}}}{{\frac{{ - 4}}{5} - \frac{{ - 3}}{5}}} = 7
\end{array}\)