Đặt \(cosx=t\)
\(\Rightarrow4t^2-2\left(\sqrt{3}-\sqrt{2}\right)t-\sqrt{6}=0\)
\(\Delta'=\left(\sqrt{3}-\sqrt{2}\right)^2+4\sqrt{6}=\left(\sqrt{3}+\sqrt{2}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{\sqrt{3}-\sqrt{2}-\sqrt{3}-\sqrt{2}}{4}=-\dfrac{\sqrt{2}}{2}\\t=\dfrac{\sqrt{3}-\sqrt{2}+\sqrt{3}+\sqrt{2}}{4}=\dfrac{\sqrt{3}}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}cosx=-\dfrac{\sqrt{2}}{2}\\cosx=\dfrac{\sqrt{3}}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\pm\dfrac{3\pi}{4}+k2\pi\\x=\pm\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)
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