\(\Leftrightarrow2cos^2x-1+2cosx-2\cdot sinx\cdot cosx+1=0\)
=>\(2\cdot cos^2x+2cosx-2sinx\cdot cosx=0\)
=>\(2\cdot cosx\cdot\left(cosx+1-sinx\right)=0\)
=>\(cosx\left(sinx-cosx-1\right)=0\)
=>\(cosx\left[\sqrt{2}\cdot sin\left(x-\dfrac{pi}{4}\right)-1\right]=0\)
=>\(\left[{}\begin{matrix}cosx=0\\sin\left(x-\dfrac{pi}{4}\right)=\dfrac{1}{\sqrt{2}}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{pi}{2}+kpi\\x-\dfrac{pi}{4}=\dfrac{pi}{4}+k2pi\\x-\dfrac{pi}{4}=\dfrac{3}{4}pi+k2pi\end{matrix}\right.\)
=>x=pi/2+kpi hoặc x=pi+k2pi
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