Q:

weird trig problem $\tan(\theta)=-\sqrt{2}\sin(\theta)$ on the interval $0 \leq \theta \leq 2\pi$

$\tan(\theta)=-\sqrt{2}\sin(\theta)$ on the interval $0 \leq \theta \lt 2\pi$
I  started off with  $[(\sin(\theta)/\cos(\theta)] \times (1/\sin(\theta) )= - \sqrt 2$,
then after simplification i got $(1/\cos(\theta))=-\sqrt 2$ and then i've got 
$$\cos(\theta)= -(\sqrt{2}/2)$$ 
and   i got $\theta = 3\pi/4,5\pi/4$ but the right answer is  $\theta=0,\pi,3\pi/4,5\pi/4$. 
I do not understand where did $0$ and $\pi$ came from?  I mean at $0$ and $\pi$ $\theta$ is not equal to $-\sqrt{2}/2$ why is then $0$ and $\pi$ included??  I would appreciate if you could explain that. 

A:

The other solutions come from solving $\sin\theta=0$. You can't just cancel the $\sin\theta$ just because it's a common factor, without considering the possibility it could be zero. 
It's like when you solve the quadratic equation $x^2-3x=0$. Would you say that the only solution is $x=3$?