Đáp án:
\[\left[ \begin{array}{l}
x = \arctan \left( { - 2} \right) + k\pi \\
x = \arctan \dfrac{{ - 2}}{3} + k\pi
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
3{\sin ^2}x + 8\sin x.\cos x + 4{\cos ^2}x = 0\,\,\,\,\left( 1 \right)\\
TH1:\,\,\,\,\cos x = 0\\
\left( 1 \right) \Leftrightarrow 3{\sin ^2}x = 0 \Leftrightarrow \sin x = 0\\
\Rightarrow {\sin ^2}x + {\cos ^2}x = 0\,\,\,\,\left( L \right)\\
TH2:\,\,\,\cos x \ne 0\\
\left( 1 \right) \Leftrightarrow \dfrac{{3{{\sin }^2}x + 8\sin x.\cos x + 4{{\cos }^2}x}}{{{{\cos }^2}x}} = 0\\
\Leftrightarrow 3.\dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} + 8.\dfrac{{\sin x}}{{\cos x}} + 4 = 0\\
\Leftrightarrow 3{\tan ^2}x + 8\tan x + 4 = 0\\
\Leftrightarrow \left( {3{{\tan }^2}x + 6\tan x} \right) + \left( {2\tan x + 4} \right) = 0\\
\Leftrightarrow 3\tan x\left( {\tan x + 2} \right) + 2.\left( {\tan x + 2} \right) = 0\\
\Leftrightarrow \left( {\tan x + 2} \right)\left( {3\tan x + 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\tan x + 2 = 0\\
3\tan x + 2 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\tan x = - 2\\
\tan x = - \dfrac{2}{3}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \arctan \left( { - 2} \right) + k\pi \\
x = \arctan \dfrac{{ - 2}}{3} + k\pi
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
