Why is $\pi/2$ omitted from the solution of $\cot x = 3 \sin 2x$? 
Why is it that the solution of 
  $$\cot x = 3 \sin 2x \quad(\text{for the interval}\; -\pi  < x < \pi)$$
  does not include $\pi/2$, even though if this is graphed, it shows intersections at $x = \pm\pi/2$?

Please see graph below. (The solutions mentioned are only four, to the exclusion of positive and negative $\pi/2$.)

Algebraically as well, one of the factors comes out to be $\cos x = 0$ (which should give $x = \pi/2$). (Hence the graph.) 


There is no reason for not including $x=\pm \pi/2$ in the intersection set. 
Surprisingly enough they have the $x=\pm \pi/2$ as solultions but they did not list them as solutions because of the quadrant confusion. 
Which quadrant are $x=\pm \pi/2??$ in??
