Question

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

Answer 1

$2sin^2x+3sin xcos x + cos^2x=0\(sin^2x+2sin xcos x+cos^2x)+(sin^2x+sin xcos 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)(2sin x+cos x)=0\begin{cases}sin x+cos x=0\2sin x+cos x=0end{cases}Rightarrowbegin{cases}sin x=-cos x\2sin x=-cos xend{cases}Rightarrowbegin{cases}tg x=-1\tgx=-frac12end{cases}\begin{cases}x=frac{3pi}4+pi n\x=arctgleft(-frac12right)+pi nend{cases},;ninmathbb{Z}$