Đáp án:

\(\begin{array}{l}
17,\\
\left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
x = \dfrac{\pi }{6} + k\pi 
\end{array} \right.\,\,\,\left( {k \in Z} \right)\\
18,\\
\left[ \begin{array}{l}
\dfrac{x}{2} = \dfrac{\pi }{6} + k2\pi \\
\dfrac{x}{2} = \dfrac{{5\pi }}{6} + k2\pi 
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{3} + k4\pi \\
x = \dfrac{{5\pi }}{3} + k4\pi 
\end{array} \right.\,\,\,\left( {k \in Z} \right)\\
19,\\
x = \dfrac{{k2\pi }}{3}\,\,\,\,\left( {k \in Z} \right)\\
20,\\
x =  - \dfrac{\pi }{4} + k\pi \,\,\,\,\left( {k \in Z} \right)\\
21,\\
\left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
x =  - \dfrac{\pi }{3} + k\pi 
\end{array} \right.\,\,\,\left( {k \in Z} \right)\\
22,\\
\left[ \begin{array}{l}
x =  - \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{7\pi }}{6} + k2\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
23,\\
\left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
x = \dfrac{\pi }{4} + k2\pi \\
x = \dfrac{{3\pi }}{4} + k2\pi 
\end{array} \right.\,\,\,\left( {k \in Z} \right)\\
24,\\
\left[ \begin{array}{l}
x = \dfrac{\pi }{3} + k\pi \\
x = \dfrac{\pi }{6} + k\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
17,\\
DKXD:\,\,\,\left\{ \begin{array}{l}
\sin x \ne 0\\
\cos x \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ne k\pi \\
x \ne \dfrac{\pi }{2} + k\pi 
\end{array} \right. \Leftrightarrow x \ne \dfrac{{k\pi }}{2}\,\,\,\,\left( {k \in Z} \right)\\
3\tan x + \sqrt 3 \cot x - 3 - \sqrt 3  = 0\\
 \Leftrightarrow 3\tan x + \dfrac{{\sqrt 3 }}{{\tan x}} - 3 - \sqrt 3  = 0\\
 \Leftrightarrow \dfrac{{3{{\tan }^2}x + \sqrt 3  + \left( { - 3 - \sqrt 3 } \right)\tan x}}{{\tan x}} = 0\\
 \Leftrightarrow 3{\tan ^2}x - \left( {3 + \sqrt 3 } \right)\tan x + \sqrt 3  = 0\\
 \Leftrightarrow \left( {3{{\tan }^2}x - 3\tan x} \right) + \left( { - \sqrt 3 \tan x + \sqrt 3 } \right) = 0\\
 \Leftrightarrow 3\tan \left( {\tan x - 1} \right) - \sqrt 3 \left( {\tan x - 1} \right) = 0\\
 \Leftrightarrow \left( {\tan x - 1} \right)\left( {3\tan x - \sqrt 3 } \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\tan x - 1 = 0\\
3\tan x - \sqrt 3  = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\tan x = 1\\
\tan x = \dfrac{{\sqrt 3 }}{3}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
x = \dfrac{\pi }{6} + k\pi 
\end{array} \right.\,\,\,\left( {k \in Z} \right)\\
18,\\
\cos x + 5\sin \dfrac{x}{2} - 3 = 0\\
 \Leftrightarrow \cos \left( {2.\dfrac{x}{2}} \right) + 5\sin \dfrac{x}{2} - 3 = 0\\
 \Leftrightarrow \left( {1 - 2{{\sin }^2}\dfrac{x}{2}} \right) + 5\sin \dfrac{x}{2} - 3 = 0\\
 \Leftrightarrow  - 2{\sin ^2}\dfrac{x}{2} + 5\sin \dfrac{x}{2} - 2 = 0\\
 \Leftrightarrow 2{\sin ^2}\dfrac{x}{2} - 5\sin \dfrac{x}{2} + 2 = 0\\
 \Leftrightarrow \left( {2\sin \dfrac{x}{2} - 1} \right)\left( {\sin \dfrac{x}{2} - 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
2\sin \dfrac{x}{2} - 1 = 0\\
\sin \dfrac{x}{2} - 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sin \dfrac{x}{2} = \dfrac{1}{2}\\
\sin \dfrac{x}{2} = 2
\end{array} \right.\\
 - 1 \le \sin \dfrac{x}{2} \le 1 \Rightarrow \sin \dfrac{x}{2} = \dfrac{1}{2}\\
 \Leftrightarrow \left[ \begin{array}{l}
\dfrac{x}{2} = \dfrac{\pi }{6} + k2\pi \\
\dfrac{x}{2} = \dfrac{{5\pi }}{6} + k2\pi 
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{3} + k4\pi \\
x = \dfrac{{5\pi }}{3} + k4\pi 
\end{array} \right.\\
19,\\
{\cos ^2}3x - \left( {1 + \sqrt 3 } \right)\cos 3x + \sqrt 3  = 0\\
 \Leftrightarrow \left( {{{\cos }^2}3x - \cos 3x} \right) + \left( { - \sqrt 3 \cos 3x + \sqrt 3 } \right) = 0\\
 \Leftrightarrow \cos 3x\left( {\cos 3x - 1} \right) - \sqrt 3 \left( {\cos 3x - 1} \right) = 0\\
 \Leftrightarrow \left( {\cos 3x - 1} \right)\left( {\cos 3x - \sqrt 3 } \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos 3x - 1 = 0\\
\cos 3x - \sqrt 3  = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\cos 3x = 1\\
\cos 3x = \sqrt 3 
\end{array} \right.\\
 - 1 \le \cos 3x \le 1 \Rightarrow \cos 3x = 1\\
\cos 3x = 1 \Leftrightarrow 3x = k2\pi  \Leftrightarrow x = \dfrac{{k2\pi }}{3}\,\,\,\,\left( {k \in Z} \right)\\
20,\\
DKXD:\,\,\,\,\sin x \ne 0 \Leftrightarrow x \ne k\pi \\
\dfrac{1}{{{{\sin }^2}x}} + 2\cot x = 0\\
 \Leftrightarrow \dfrac{1}{{{{\sin }^2}x}} + 2.\dfrac{{\cos x}}{{\sin x}} = 0\\
 \Leftrightarrow \dfrac{{1 + 2\sin x.\cos x}}{{{{\sin }^2}x}} = 0\\
 \Leftrightarrow 1 + 2\sin x.\cos x = 0\\
 \Leftrightarrow 1 + \sin 2x = 0\\
 \Leftrightarrow \sin 2x =  - 1\\
 \Leftrightarrow 2x =  - \dfrac{\pi }{2} + k2\pi \\
 \Leftrightarrow x =  - \dfrac{\pi }{4} + k\pi \,\,\,\,\left( {k \in Z} \right)
\end{array}\)

\(\begin{array}{l}
21,\\
DKXD:\,\,\,\cos x \ne 0 \Leftrightarrow x \ne \dfrac{\pi }{2} + k\pi \,\,\,\,\left( {k \in Z} \right)\\
{\tan ^2}x + \left( {\sqrt 3  - 1} \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)\\
22,\\
2{\cos ^2}x + 5\sin x + 1 = 0\\
 \Leftrightarrow 2.\left( {1 - {{\sin }^2}x} \right) + 5\sin x + 1 = 0\\
 \Leftrightarrow 2 - 2{\sin ^2}x + 5\sin x + 1 = 0\\
 \Leftrightarrow  - 2{\sin ^2}x + 5\sin x + 3 = 0\\
 \Leftrightarrow 2{\sin ^2}x - 5\sin x - 3 = 0\\
 \Leftrightarrow \left( {2\sin x + 1} \right)\left( {\sin x - 3} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
2\sin x + 1 = 0\\
\sin x - 3 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sin x =  - \dfrac{1}{2}\\
\sin x = 3
\end{array} \right.\\
 - 1 \le \sin x \le 1 \Rightarrow \sin x =  - \dfrac{1}{2}\\
\sin x =  - \dfrac{1}{2} \Leftrightarrow \left[ \begin{array}{l}
x =  - \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{7\pi }}{6} + k2\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
23,\\
4{\cos ^3}x + 3\sqrt 2 \sin 2x = 8\cos x\\
 \Leftrightarrow 4{\cos ^3}x + 3\sqrt 2 .2\sin x.\cos x - 8\cos x = 0\\
 \Leftrightarrow 4{\cos ^3}x + 6\sqrt 2 \sin x.\cos x - 8\cos x = 0\\
 \Leftrightarrow 2\cos x\left( {2{{\cos }^2}x + 3\sqrt 2 \sin x - 4} \right) = 0\\
 \Leftrightarrow \cos x.\left[ {2.\left( {1 - {{\sin }^2}x} \right) + 3\sqrt 2 \sin x - 4} \right] = 0\\
 \Leftrightarrow \cos x.\left( {2 - 2{{\sin }^2}x + 3\sqrt 2 \sin x - 4} \right) = 0\\
 \Leftrightarrow \cos x.\left( { - 2{{\sin }^2}x + 3\sqrt 2 \sin x - 2} \right) = 0\\
 \Leftrightarrow \cos x.\left( {2{{\sin }^2}x - 3\sqrt 2 \sin x + 2} \right) = 0\\
 \Leftrightarrow \cos x.\left( {2\sin x - \sqrt 2 } \right)\left( {\sin x - \sqrt 2 } \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
2\sin x - \sqrt 2  = 0\\
\sin x - \sqrt 2  = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
\sin x = \dfrac{{\sqrt 2 }}{2}\\
\sin x = \sqrt 2 
\end{array} \right.\\
 - 1 \le \sin x \le 1 \Rightarrow \left[ \begin{array}{l}
\cos x = 0\\
\sin x = \dfrac{{\sqrt 2 }}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
x = \dfrac{\pi }{4} + k2\pi \\
x = \dfrac{{3\pi }}{4} + k2\pi 
\end{array} \right.\,\,\,\left( {k \in Z} \right)\\
24,\\
DKXD:\,\,\,\cos x \ne 0 \Leftrightarrow x \ne \dfrac{\pi }{2} + k\pi \\
\dfrac{3}{{{{\cos }^2}x}} + 2\sqrt 3 \tan x = 6\\
 \Leftrightarrow \dfrac{3}{{{{\cos }^2}x}} + 2\sqrt 3 .\dfrac{{\sin x}}{{\cos x}} = 6\\
 \Leftrightarrow \dfrac{{3 + 2\sqrt 3 \sin x.\cos x}}{{{{\cos }^2}x}} = 6\\
 \Leftrightarrow 3 + 2\sqrt 3 \sin x.\cos x = 6{\cos ^2}x\\
 \Leftrightarrow 3.\left( {{{\sin }^2}x + {{\cos }^2}x} \right) + 2\sqrt 3 \sin x.\cos x = 6{\cos ^2}x\\
 \Leftrightarrow 3{\sin ^2}x + 3{\cos ^2}x + 2\sqrt 3 \sin x.\cos x = 6{\cos ^2}x\\
 \Leftrightarrow 3{\sin ^2}x + 2\sqrt 3 \sin x.\cos x - 3{\cos ^2}x = 0\\
 \Leftrightarrow \left( {3{{\sin }^2}x + 3\sqrt 3 \sin x.\cos x} \right) + \left( { - \sqrt 3 \sin x.\cos x - 3{{\cos }^2}x} \right) = 0\\
 \Leftrightarrow 3\sin x.\left( {\sin x + \sqrt 3 \cos x} \right) - \sqrt 3 \cos x.\left( {\sin x + \sqrt 3 \cos x} \right) = 0\\
 \Leftrightarrow \left( {\sin x + \sqrt 3 \cos x} \right)\left( {3\sin x - \sqrt 3 \cos x} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\sin x + \sqrt 3 \cos x = 0\\
3\sin x - \sqrt 3 \cos x = 0
\end{array} \right.\\
\cos x \ne 0 \Rightarrow \left[ \begin{array}{l}
\dfrac{{\sin x}}{{\cos x}} - \sqrt 3  = 0\\
3.\dfrac{{\sin x}}{{\cos x}} - \sqrt 3  = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
\tan x - \sqrt 3  = 0\\
3\tan x - \sqrt 3  = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\tan x = \sqrt 3 \\
\tan x = \dfrac{{\sqrt 3 }}{3}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{3} + k\pi \\
x = \dfrac{\pi }{6} + k\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)