It is given $(\tan \theta +1)(\sin^² \theta -3\cos^² \theta)=0$, find the possible values of $\tan \theta$ 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.
 A: 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.
A: 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 $n\in\mathbb{Z}$
