2 sin^2x+3sin x cos x + cos^2x=0

Решите задачу:

$2\sin^2x+3\sin x\cos x + \cos^2x=0\\(\sin^2x+2\sin x\cos x+\cos^2x)+(\sin^2x+\sin x\cos x)=0\\\sin x+\cos x)^2+\sin x(\sin x+\cos x)=0\\(\sin x+\cos x)(\sin x+\cos x+sin x)=0\\(\sin x+\cos x)(2\sin x+\cos x)=0\\\begin{cases}\sin x+\cos x=0\\2\sin x+\cos x=0\end{cases}\Rightarrow\begin{cases}\sin x=-\cos x\\2\sin x=-\cos x\end{cases}\Rightarrow\begin{cases}tg x=-1\\tgx=-\frac12\end{cases}\\\begin{cases}x=\frac{3\pi}4+\pi n\\x=arctg\left(-\frac12\right)+\pi n\end{cases},\;n\in\mathbb{Z}$