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Đặt \(tanx=t\Rightarrow cos2x=\frac{1-t^2}{1+t^2}\)
Pt trở thành: \(\frac{2-2t^2}{1+t^2}+t=\frac{4}{5}\Leftrightarrow5t^3-14t^2+5t+6=0\)
\(\Leftrightarrow\left(t-2\right)\left(5t^2-4t-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}t=2\\t=\frac{2+\sqrt{19}}{5}\\t=\frac{2-\sqrt{19}}{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=arctan\left(2\right)+k\pi\\x=arctan\left(\frac{2+\sqrt{19}}{5}\right)+k\pi\\x=arctan\left(\frac{2-\sqrt{19}}{5}\right)+k\pi\end{matrix}\right.\)
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