Have You Got Them All?

Relevant wiki: Trigonometric Equations - Problem Solving - Medium

A lot of people are being tricked by this problem, so I will explain what they are doing incorrectly. BTW, this was one of the problems in a 2015 Maths test for Grade 12 in my school, and none of them got it right.

The starting is fairly standard. Either one of the two given brackets must be 0. If $\tan^2 \theta - 3 = 0$ , $\tan \theta = \pm \sqrt{3}$ , so we get solutions $\theta = \frac {\pi}{3}, \frac {2\pi}{3}, \frac {4\pi}{3}, \frac {5\pi}{3}$ , which are all possible solutions. If we have $\sin{\theta}+1=0$ , we get $\sin \theta = -1$ , so $\theta = \frac {3\pi}{2}$ . (Here comes the trick) However, if $\theta=\frac {3\pi}{2}$ , $\tan^2 \theta$ is infinity, and the product of 0 and infinity is undefined, so this cannot be a solution.

Therefore, there are only 4 satisfying solutions.

We graph the function between $0\ \ and\ \ 2\pi$ and we get four values only cosponsoring to $Tan^2\theta -3 =0,\ but Sin\theta + 1=0,\ makes\ Tan\theta^2-3= \ \infty. \ And\ 0* \infty\ not\ defined.$ .