[tex]2\cos(x)\sin(x)=\cos(x)\ \ \ \ |\text{subtract}\ \cos(x)\ \text{from both sides}\\\\2\cos(x)\sin(x)-\cos(x)=0\\\\\cos(x)(2\sin(x)-1)=0\iff\cos(x)=0\ \vee\ 2\sin(x)-1=0\\\\\cos(x)=0\to x=\dfrac{\pi}{2}+k\pi,\ k\in\mathbb{Z}\\\\2\sin(x)-1=0\qquad|\text{add 1 to both sides}\\\\2\sin(x)=1\qquad|\text{divide both sides by 2}\\\\\sin(x)=\dfrac{1}{2}\to x=\dfrac{\pi}{6}+2k\pi\ \vee\ x=\dfrac{5\pi}{6}+2k\pi,\ k\in\mathbb{Z}[/tex]
[tex]Answer:\\\\x=\dfrac{\pi}{2}+k\pi\ \vee\ x=\dfrac{\pi}{6}+2k\pi\ \vee\ x=\dfrac{5\pi}{6}+2k\pi\ for\ k\in\mathbb{Z}[/tex]
