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Not sure if this type of question is allowed, I need to know if my working and my answer is correct. The working is kind of similar to solving a quadratic

$(\tan \theta+1)(\sin^2 \theta -3 \cos^2 \theta)=0$

$(\tan \theta +1)(\sin^2 \theta-3\cos^2 \theta)/(\sin^2 \theta -3\cos^2 \theta )=0/(\sin^2 \theta -3\cos^2 \theta )$

$tanθ+1=0$

first possible value

$tanθ=-1$

$(tanθ+1)(sin²θ-3cos²θ)=0$

$(tanθ+1)(sin²θ-3cos²θ)/(tanθ+1)=0/(tanθ+1)$

$(sin²θ-3cos²θ)=0$

$sin²θ=3cos²θ$

$(sin²θ)/(cos²θ)=3$

since $(sin²θ)/(cos²θ)=tan²θ$

$tan²θ=3$

second and third possible values

$tanθ=±√3$

This kind of confused me that there was 3 possible values so I wasn't sure.

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    $\begingroup$ Your conclusions (the three values) are correct. I would only suggest that you can/should avoid the steps where you divide by factors. Being given that $$(\tan\theta + 1 )(\sin^2\theta-3\cos^2\theta) = 0$$ just jump to saying that $$\tan\theta + 1 = 0 \quad\text{OR}\quad \sin^2\theta-3\cos^2\theta = 0$$ and solve each case (as you did). The jump is justified by the fact that $ab=0$ if and only if either $a=0$ or $b=0$ (or both). $\endgroup$ – Blue Feb 18 '15 at 1:58
  • $\begingroup$ you are correct. can you please put dollar symbol around the mathematical expressions. $\endgroup$ – abel Feb 18 '15 at 1:59
  • $\begingroup$ you can look at the first two lines i edited and can you edit the rest of your question. $\endgroup$ – abel Feb 18 '15 at 2:18
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I am not exactly sure why people are voting to close this, but yes, proof verification is allowed on this site and you have clearly put in the work! Your answer and work looks perfectly fine to me. As for explaining why you have three solutions, that is because you can view this equation as a cubic in terms of tangent, and cubics can have up to three roots. Notice that $$(\tan(\theta)+1)\left(\sin^2(\theta)-3\cos^2(\theta)\right) = (\tan(\theta)+1)\left(\tan^2(\theta)-3\right)\cos^2(\theta)$$ so you want $$(\tan(\theta)+1)\left(\tan^2(\theta)-3\right)\cos^2(\theta)=0$$ You already know $\theta \neq \frac{\pi}{2}+k\pi$, so you can divide by the nonzero $\cos^2(\theta)$ to get $$(\tan(\theta)+1)\left(\tan^2(\theta)-3\right)=0$$ which is a cubic in terms of tangent, now factored in a way that makes the values of $\tan(\theta)$ very clear.

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  • $\begingroup$ but you won't even bother to edit the question. $\endgroup$ – abel Feb 18 '15 at 2:08
  • $\begingroup$ @abel I mean.. am I supposed to? It's not the most beautiful formatting I've seen but it's loads better than some train wreck questions that show up on here. $\endgroup$ – graydad Feb 18 '15 at 2:10
  • $\begingroup$ i don't think there is anything supposed. i think it is a courtesy to edit what you can to make it better. my latex skill is not very good. i try to make the questions reasonable. i did not mean to be critical. $\endgroup$ – abel Feb 18 '15 at 2:14
  • $\begingroup$ @abel nah you are fine. And the edits you made look good! I'd probably argue that now Thomas Winkworth can you at what you did and edit the rest of his question accordingly. $\endgroup$ – graydad Feb 18 '15 at 2:16
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If $(2\cos(2\theta)+1)(\tan(\theta)+1)=0$ then

$\cos(2\theta)=-1/2$ implies $\theta=\pi n+\pi/3$ and $\theta=\pi n-\pi/3$

ó

$\tan(\theta)=-1$ implies $\theta=\pi n-\pi/4$ with $n\in\mathbb{Z}$

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