\[\begin{array}{l}
\dfrac{1}{{\sin x}} + \tan x = \dfrac{{12}}{5} \Leftrightarrow \dfrac{1}{{\sin x}} + \dfrac{{\sin x}}{{\cos x}} = \dfrac{{12}}{5} \Leftrightarrow \dfrac{{{{\sin }^2}x + 1}}{{\sin x\cos x}} = \dfrac{{12}}{5}\\
\Leftrightarrow 2({\sin ^2}x + 1) = \dfrac{{12}}{5}\sin 2x \Leftrightarrow 1 - \cos 2x + 2 = \dfrac{{12}}{5}\sin 2x\\
\Leftrightarrow \sin 2x + \cos 2x = \dfrac{4}{5} \Leftrightarrow \sqrt 2 \sin \left( {2x + \dfrac{\pi }{4}} \right) = \dfrac{4}{5} \Leftrightarrow \sin \left( {2x + \dfrac{\pi }{4}} \right) = \dfrac{{2\sqrt 2 }}{5}\\
\Rightarrow \left[ \begin{array}{l}
2x + \dfrac{\pi }{4} = \arcsin \left( {\dfrac{{2\sqrt 2 }}{5}} \right)\\
2x + \dfrac{\pi }{4} = \pi - \arcsin \left( {\dfrac{{2\sqrt 2 }}{5}} \right)
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = \dfrac{1}{2}\arcsin \left( {\dfrac{{2\sqrt 2 }}{5}} \right) - \dfrac{\pi }{8} + k2\pi \\
x = \dfrac{1}{2}\arcsin \left( {\dfrac{{2\sqrt 2 }}{5}} \right) + \dfrac{\pi }{8} + k2\pi
\end{array} \right.
\end{array}\]
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