Đáp án:
\[\left[ \begin{array}{l}
x = \arctan \left( {4 + 2\sqrt 3 } \right) + k\pi \\
x = \arctan \left( {4 - 2\sqrt 3 } \right) + k\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{\sin ^2}x - 8\sin x.\cos x + 4{\cos ^2}x = 0\,\,\,\,\left( 1 \right)\\
TH1:\,\,\,\cos x = 0\\
\left( 1 \right) \Leftrightarrow {\sin ^2}x = 0 \Leftrightarrow \sin x = 0\\
\Rightarrow {\sin ^2}x + {\cos ^2}x = 0\,\,\,\,\,\left( L \right)\\
TH2:\,\,\,\,\cos x \ne 0\\
\left( 1 \right) \Leftrightarrow \dfrac{{{{\sin }^2}x - 8\sin x.\cos x + 4{{\cos }^2}x}}{{{{\cos }^2}x}} = 0\\
\Leftrightarrow \dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} - 8.\dfrac{{\sin x}}{{\cos x}} + 4 = 0\\
\Leftrightarrow {\tan ^2}x - 8\tan x + 4 = 0\\
\Leftrightarrow {\tan ^2}x - 8\tan x + 16 = 12\\
\Leftrightarrow {\left( {\tan x - 4} \right)^2} = 12\\
\Leftrightarrow \left[ \begin{array}{l}
\tan x - 4 = 2\sqrt 3 \\
\tan x - 4 = - 2\sqrt 3
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\tan x = 4 + 2\sqrt 3 \\
\tan x = 4 - 2\sqrt 3
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \arctan \left( {4 + 2\sqrt 3 } \right) + k\pi \\
x = \arctan \left( {4 - 2\sqrt 3 } \right) + k\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
