Verified Answer :

\(2sinxcosx+cosx=0\\\\cosx(2sinx+1)=0\iff cosx=0\ \vee\ 2sinx+1=0\\\\cosx=0\ \vee\ 2sinx=-1\ /:2\\\\cosx=0\ \vee\ sinx=-\frac{1}{2}\\\\x=\frac{\pi}{2}+k\pi\ \vee\ x=-\frac{\pi}{6}+2k\pi\ \vee\ x=\frac{7\pi}{6}+2k\pi\ to\ k\in\mathbb{Z}\)

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