Đáp án:
\(\left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
x = \arccos \dfrac{3}{{\sqrt {10} }} + k\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\)
Giải thích các bước giải:
ĐKXĐ: \(\cos x \ne 0 \Leftrightarrow x \ne \dfrac{\pi }{2} + k\pi \)
Ta có:
\(\begin{array}{l}
2\cos x + 4\sin x = \dfrac{3}{{\cos x}}\\
\Leftrightarrow 2.{\cos ^2}x + 4\sin x.\cos x = 3\\
\Leftrightarrow 2{\cos ^2}x + 4\sin x.\cos x = 3.\left( {{{\sin }^2}x + {{\cos }^2}x} \right)\\
\Leftrightarrow 2{\cos ^2}x + 4\sin x.\cos x = 3\sin {x^2} + 3{\cos ^2}x\\
\Leftrightarrow 3{\sin ^2}x - 4\sin x.\cos x + {\cos ^2}x = 0\\
\Leftrightarrow \left( {3{{\sin }^2}x - 3\sin x.\cos x} \right) + \left( { - \sin x.\cos x + {{\cos }^2}x} \right) = 0\\
\Leftrightarrow 3\sin x.\left( {\sin x - \cos x} \right) - \cos x.\left( {\sin x - \cos x} \right) = 0\\
\Leftrightarrow \left( {\sin x - \cos x} \right).\left( {3\sin x - \cos x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x - \cos x = 0\\
3\sin x - \cos x = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{1}{{\sqrt 2 }}.\sin x - \dfrac{1}{{\sqrt 2 }}.\cos x = 0\\
\dfrac{3}{{\sqrt {10} }}\sin x - \dfrac{1}{{\sqrt {10} }}.\cos x = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x.\cos \dfrac{\pi }{4} - \cos x.\sin \dfrac{\pi }{4} = 0\\
\sin x.\cos \left( {\arccos \dfrac{3}{{\sqrt {10} }}} \right) - \cos x.\sin \left( {\arccos \dfrac{3}{{\sqrt {10} }}} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\sin \left( {x - \dfrac{\pi }{4}} \right) = 0\\
\sin \left( {x - \arccos \dfrac{3}{{\sqrt {10} }}} \right) = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x - \dfrac{\pi }{4} = k\pi \\
x - \arccos \dfrac{3}{{\sqrt {10} }} = k\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
x = \arccos \dfrac{3}{{\sqrt {10} }} + k\pi
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)
