# 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: $$\cos^2x\left(\tan^2x+3\tan x+2\right)=0.$$

• Here's the MathJax tutorial. – SarGe Jun 19 '20 at 14:14
• Is it supposed to be equal to zero? I do not see an equation. – Vasya Jun 19 '20 at 14:15
• You will need the $$tan(x/2)$$ substitution. – Dr. Sonnhard Graubner Jun 19 '20 at 14:19
• If you check the equality for either $x=\frac\pi2$ or $x=\frac{3\pi}2$, you would end up with $1=0$. – user170231 Jun 19 '20 at 14:20
• Although $x=\pi/2$ and $x=3\pi/2$ make the $\cos^2x$ factor vanish, they make the $\tan^2x+\cdots$ factor undefined, which is a problem. :) Testing $x=\pi/2$ and $x=3\pi/2$ in the original equation shows that they are not solutions. – Blue Jun 19 '20 at 14:21

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