Respuesta :

[tex]2\sin x\cos x=\cos x\\\\2\sin x\cos x-\cos x=0\\\\\cos x(2\sin x-1)=0\iff\ \cos x=0\ or\ 2\sin x-1=0\\\\\cos x=0\ if\ x=\dfrac{\pi}{2}+k\pi\\\\2\sin x-1=0\\\\2\sin x=1\\\\\sin x=\dfrac{1}{2}\ if\ x=\dfrac{\pi}{6}+2k\pi\ or\ x=\dfrac{5\pi}{6}+2k\pi\\\\Answer:\boxed{x=\dfrac{\pi}{2}+k\pi\ or\ x=\dfrac{\pi}{6}+2k\pi\ or\ x=\dfrac{5\pi}{6}+2k\pi\ /k\in\mathbb{Z}/}[/tex]
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