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:$$\cos^2x\left(\tan^2x+3\tan x+2\right)=0.$$
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$\begingroup$Look at this $$x^2+3xy+2y^2=(x+y)(x+2y)$$ can you see??
$\endgroup$ 1 $\begingroup$Use that $$\sin(x)=2\,{\frac {\tan \left( x/2 \right) }{1+ \left( \tan \left( x/2 \right) \right) ^{2}}} $$ and$$\cos(x)={\frac {1- \left( \tan \left( x/2 \right) \right) ^{2}}{1+ \left( \tan \left( x/2 \right) \right) ^{2}}} $$
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