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