Đáp án:

\(\begin{array}{l}
a,\\
\left[ \begin{array}{l}
x = \dfrac{\pi }{3} + k2\pi \\
x =  - \dfrac{\pi }{3} + k2\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
b,\\
\left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
x = \dfrac{\pi }{3} + k\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
c,\\
\left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
x = \arctan 4 + k\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
a,\\
6{\sin ^2}x - 5\cos x - 2 = 0\\
 \Leftrightarrow 6.\left( {1 - {{\cos }^2}x} \right) - 5\cos x - 2 = 0\\
 \Leftrightarrow 6 - 6{\cos ^2}x - 5\cos x - 2 = 0\\
 \Leftrightarrow  - 6{\cos ^2}x - 5\cos x + 4 = 0\\
 \Leftrightarrow 6{\cos ^2}x + 5\cos x - 4 = 0\\
 \Leftrightarrow \left( {2\cos x - 1} \right)\left( {3\cos x + 4} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
2\cos x - 1 = 0\\
3\cos x + 4 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\cos x = \dfrac{1}{2}\\
\cos x =  - \dfrac{4}{3}
\end{array} \right.\\
 - 1 \le \cos x \le 1 \Rightarrow \cos x = \dfrac{1}{2}\\
\cos x = \dfrac{1}{2}\\
 \Leftrightarrow \cos x = \cos \dfrac{\pi }{3}\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{3} + k2\pi \\
x =  - \dfrac{\pi }{3} + k2\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
b,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
\sin x \ne 0\\
\cos x \ne 0
\end{array} \right. \Leftrightarrow x \ne \dfrac{{k\pi }}{2}\,\,\,\,\left( {k \in Z} \right)\\
\tan x + \sqrt 3 \cot x - 1 - \sqrt 3  = 0\\
 \Leftrightarrow \tan x + \dfrac{{\sqrt 3 }}{{\tan x}} - 1 - \sqrt 3  = 0\\
 \Leftrightarrow \dfrac{{{{\tan }^2}x + \sqrt 3  + \left( { - 1 - \sqrt 3 } \right)\tan x}}{{\tan x}} = 0\\
 \Leftrightarrow {\tan ^2}x + \left( { - 1 - \sqrt 3 } \right)\tan x + \sqrt 3  = 0\\
 \Leftrightarrow \left( {{{\tan }^2}x - \tan x} \right) + \left( { - \sqrt 3 \tan x + \sqrt 3 } \right) = 0\\
 \Leftrightarrow \tan x\left( {\tan x - 1} \right) - \sqrt 3 \left( {\tan x - 1} \right) = 0\\
 \Leftrightarrow \left( {\tan x - 1} \right)\left( {\tan x - \sqrt 3 } \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\tan x - 1 = 0\\
\tan x - \sqrt 3  = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\tan x = 1\\
\tan x = \sqrt 3 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
x = \dfrac{\pi }{3} + k\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
c,\\
{\sin ^2}x + 5\sin x.\cos x - 2{\cos ^2}x = 2\\
 \Leftrightarrow {\sin ^2}x + 5\sin x.\cos x - 2{\cos ^2}x = 2\left( {{{\sin }^2}x + {{\cos }^2}x} \right)\\
 \Leftrightarrow {\sin ^2}x + 5\sin x.\cos x - 2{\cos ^2}x = 2{\sin ^2}x + 2{\cos ^2}x\\
 \Leftrightarrow {\sin ^2}x - 5\sin x.\cos x + 4{\cos ^2}x = 0\,\,\,\,\,\left( 1 \right)\\
TH1:\,\,\,\cos x = 0\\
\left( 1 \right) \Leftrightarrow {\sin ^2}x = 0 \Rightarrow \sin x = 0\\
 \Rightarrow {\sin ^2}x + {\cos ^2}x = 0\,\,\,\,\left( L \right)\\
TH2:\,\,\,\cos x \ne 0\\
\left( 1 \right) \Leftrightarrow \dfrac{{{{\sin }^2}x - 5\sin x.\cos x + 4{{\cos }^2}x}}{{{{\cos }^2}x}} = 0\\
 \Leftrightarrow \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} - 5.\dfrac{{\sin x}}{{\cos x}} + 4 = 0\\
 \Leftrightarrow {\tan ^2}x - 5\tan x + 4 = 0\\
 \Leftrightarrow \left( {\tan x - 1} \right)\left( {\tan x - 4} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\tan x - 1 = 0\\
\tan x - 4 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\tan x = 1\\
\tan x = 4
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
x = \arctan 4 + k\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)