Solving $\sin^2x + 3\sin x\cos x + 2\cos^2x=0$, for $0\leq x\leq 2\pi$ 
Solve
$$\sin^2x + 3\sin x\cos x + 2\cos^2x=0$$
for $0\leq x\leq 2\pi$.

My answers are
$$x=2.03, 5.18 \qquad\text{or}\qquad
x=\frac{3\pi}{4},\frac{7\pi}{4} \qquad\text{or}\qquad
x=\frac{\pi}{2}, \frac{3\pi}{2},$$
but the answer states $x=2.03, 5.18$ or $x=3\pi/4,7\pi/4$ only.
I got $x=\pi/2, 3\pi/2$ from $(\cos x)^2=0$,
where it is a factor in one of my steps:
$$\cos^2x\left(\tan^2x+3\tan x+2\right)=0.$$

Use that $$\sin(x)=2\,{\frac {\tan \left( x/2 \right) }{1+ \left( \tan \left( x/2
 \right)  \right) ^{2}}}
$$ and
$$\cos(x)={\frac {1- \left( \tan \left( x/2 \right)  \right) ^{2}}{1+ \left( 
\tan \left( x/2 \right)  \right) ^{2}}}
 $$
