Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
2{\cos ^2}x - 3\cos x + 1 = 0\\
\Leftrightarrow \left( {2{{\cos }^2}x - 2\cos x} \right) + \left( { - \cos x + 1} \right) = 0\\
\Leftrightarrow 2\cos x.\left( {\cos x - 1} \right) - \left( {\cos x - 1} \right) = 0\\
\Leftrightarrow \left( {\cos x - 1} \right)\left( {2\cos x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x - 1 = 0\\
2\cos x - 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = 1\\
\cos x = \dfrac{1}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = k2\pi \\
x = \pm \dfrac{\pi }{3} + k2\pi
\end{array} \right.\,\,\,\,\,\,\,\left( {k \in Z} \right)\\
b,\\
5{\sin ^2}x - 4\sin x - 1 = 0\\
\Leftrightarrow \left( {5{{\sin }^2}x - 5\sin x} \right) + \left( {\sin x - 1} \right) = 0\\
\Leftrightarrow 5\sin x.\left( {\sin x - 1} \right) + \left( {\sin x - 1} \right) = 0\\
\Leftrightarrow \left( {\sin x - 1} \right)\left( {5\sin x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x - 1 = 0\\
5\sin x + 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = 1\\
\sin x = - \dfrac{1}{5}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k2\pi \\
x = \arcsin \left( { - \dfrac{1}{5}} \right) + k2\pi \\
x = \pi - \arcsin \left( { - \dfrac{1}{5}} \right) + k2\pi
\end{array} \right.\,\,\,\,\,\,\left( {k \in Z} \right)\\
c,\\
\cos 2x - 3\cos x - 4 = 0\\
\Leftrightarrow \left( {2{{\cos }^2}x - 1} \right) - 3\cos x - 4 = 0\\
\Leftrightarrow 2{\cos ^2}x - 3\cos x - 5 = 0\\
\Leftrightarrow \left( {2{{\cos }^2}x + 2\cos x} \right) + \left( { - 5\cos x - 5} \right) = 0\\
\Leftrightarrow 2\cos x.\left( {\cos x + 1} \right) - 5.\left( {\cos x + 1} \right) = 0\\
\Leftrightarrow \left( {\cos x + 1} \right)\left( {2\cos x - 5} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x + 1 = 0\\
2\cos x - 5 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = - 1\\
\cos x = \dfrac{5}{2}
\end{array} \right.\\
- 1 \le \cos x \le 1 \Rightarrow \cos x = - 1 \Leftrightarrow x = \pi + k2\pi \,\,\,\,\,\left( {k \in Z} \right)\\
d,\\
\tan x + 2\cot x - 3 = 0\\
\Leftrightarrow \tan x + \dfrac{2}{{\tan x}} - 3 = 0\\
\Leftrightarrow \dfrac{{{{\tan }^2}x + 2 - 3\tan x}}{{\tan x}} = 0\\
\Leftrightarrow {\tan ^2}x - 3\tan x + 2 = 0\\
\Leftrightarrow \left( {\tan x - 1} \right)\left( {\tan x - 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\tan x - 1 = 0\\
\tan x - 2 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\tan x = 1\\
\tan x = 2
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
x = \arctan 2 + k\pi
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\\
e,\\
4{\sin ^2}2x + 8{\cos ^2}x - 9 = 0\\
\Leftrightarrow 4.{\left( {2\sin x.\cos x} \right)^2} + 8{\cos ^2}x - 9 = 0\\
\Leftrightarrow 16.{\sin ^2}x.{\cos ^2}x + 8{\cos ^2}x - 9 = 0\\
\Leftrightarrow 16.\left( {1 - {{\cos }^2}x} \right).{\cos ^2}x + 8{\cos ^2}x - 9 = 0\\
\Leftrightarrow 16{\cos ^2}x - 16{\cos ^4}x + 8{\cos ^2}x - 9 = 0\\
\Leftrightarrow - 16{\cos ^4}x + 24{\cos ^2}x - 9 = 0\\
\Leftrightarrow 16{\cos ^4}x - 24{\cos ^2}x + 9 = 0\\
\Leftrightarrow {\left( {4{{\cos }^2}x} \right)^2} - 2.4{\cos ^2}x.3 + {3^2} = 0\\
\Leftrightarrow {\left( {4{{\cos }^2}x - 3} \right)^2} = 0\\
\Leftrightarrow 4{\cos ^2}x - 3 = 0\\
\Leftrightarrow {\cos ^2}x = \dfrac{3}{4}\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = \dfrac{{\sqrt 3 }}{2}\\
\cos x = - \dfrac{{\sqrt 3 }}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \pm \dfrac{\pi }{6} + k2\pi \\
x = \pm \dfrac{{5\pi }}{6} + k2\pi
\end{array} \right.\,\,\,\,\,\,\left( {k \in Z} \right)\\
f,\\
\cos 2x + 2\cos x - 3 = 2{\sin ^2}\dfrac{x}{2}\\
\Leftrightarrow \left( {2{{\cos }^2}x - 1} \right) + 2\cos x - 4 = 2{\sin ^2}\dfrac{x}{2} - 1\\
\Leftrightarrow 2{\cos ^2}x + 2\cos x - 5 = - \cos x\\
\Leftrightarrow 2{\cos ^2}x + 3\cos x - 5 = 0\\
\Leftrightarrow \left( {2{{\cos }^2}x - 2\cos x} \right) + \left( {5\cos x - 5} \right) = 0\\
\Leftrightarrow 2\cos x.\left( {\cos x - 1} \right) + 5.\left( {\cos x - 1} \right) = 0\\
\Leftrightarrow \left( {\cos x - 1} \right)\left( {2\cos x + 5} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x - 1 = 0\\
2\cos x + 5 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\cos x = 1\\
\cos x = - \dfrac{5}{2}
\end{array} \right.\\
- 1 \le \cos x \le 1 \Rightarrow \cos x = 1 \Leftrightarrow x = k2\pi \,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
