\(\left|sinx\right|=cos2x\)
=>\(\left\{{}\begin{matrix}cos2x>0\\cos^22x=sin^2x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{4}pi+kpi< x< \dfrac{5}{4}pi+kpi\\\left(cos2x-sinx\right)\left(cos2x+sinx\right)=0\end{matrix}\right.\)
(cos2x-sinx)(cos2x+sinx)=0
=>cos2x=sin x hoặc cos2x=-sin x=sin(-x)
=>cos2x=cos(pi/2-x) hoặc cos2x=cos(pi/2+x)
=>2x=pi/2-x+k2pi hoặc 2x=-pi/2+x+k2pi hoặc 2x=pi/2+x+k2pi hoặc 2x=-pi/2-x+k2pi
=>x=pi/6+k2pi/3 hoặc x=-pi/2+k2pi hoặc x=pi/2+k2pi hoặc x=-pi/6+k2pi/3
=>\(\left[{}\begin{matrix}x=\pm\dfrac{pi}{6}+\dfrac{k2pi}{3}\\x=\pm\dfrac{pi}{2}+k2pi\end{matrix}\right.\)
mà \(\dfrac{3}{4}pi+kpi< x< \dfrac{5}{4}pi+kpi\)
nên \(x=\dfrac{5}{6}pi+k2pi;x=\dfrac{7}{6}pi+k2pi\)
