ĐK: \(x\ne k\pi;x\ne\dfrac{2\pi}{3}+k2\pi\)
\(\dfrac{\left(1-2cosx\right)\left(1+cosx\right)}{\left(1+2cosx\right)sinx}=1\)
\(\Leftrightarrow\left(1-2cosx\right)\left(1+cosx\right)=\left(1+2cosx\right)sinx\)
\(\Leftrightarrow1-2cosx+cosx-2cos^2x=sinx+2sinx.cosx\)
\(\Leftrightarrow sin2x+cos2x+sinx+cosx=0\)
\(\Leftrightarrow2sin\dfrac{3x}{2}.cos\dfrac{x}{2}+2cos\dfrac{3x}{2}.cos\dfrac{x}{2}=0\)
\(\Leftrightarrow2cos\dfrac{x}{2}\left(sin\dfrac{3x}{2}+cos\dfrac{3x}{2}\right)=0\)
\(\Leftrightarrow2\sqrt{2}cos\dfrac{x}{2}.sin\left(\dfrac{3x}{2}+\dfrac{\pi}{4}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos\dfrac{x}{2}=0\\sin\left(\dfrac{3x}{2}+\dfrac{\pi}{4}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{2}=\dfrac{\pi}{2}+k\pi\\\dfrac{3x}{2}+\dfrac{\pi}{4}=k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\pi+k2\pi\\x=-\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\end{matrix}\right.\)
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