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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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    $\begingroup$ Here's the MathJax tutorial. $\endgroup$ – SarGe Jun 19 '20 at 14:14
  • $\begingroup$ Is it supposed to be equal to zero? I do not see an equation. $\endgroup$ – Vasya Jun 19 '20 at 14:15
  • $\begingroup$ You will need the $$tan(x/2)$$ substitution. $\endgroup$ – Dr. Sonnhard Graubner Jun 19 '20 at 14:19
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    $\begingroup$ If you check the equality for either $x=\frac\pi2$ or $x=\frac{3\pi}2$, you would end up with $1=0$. $\endgroup$ – user170231 Jun 19 '20 at 14:20
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    $\begingroup$ 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. $\endgroup$ – Blue Jun 19 '20 at 14:21
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Look at this $$x^2+3xy+2y^2=(x+y)(x+2y)$$ can you see??

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  • $\begingroup$ A nice idea to solve this equation $\endgroup$ – Dr. Sonnhard Graubner Jun 19 '20 at 14:24
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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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  • $\begingroup$ You can use this also in other situations. $\endgroup$ – Dr. Sonnhard Graubner Jun 19 '20 at 14:27

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