Solve the following trigonometric equation using product cancellation Say I want to solve this equation using product cancellation:
$$\cos(2x+\frac {\pi}{2})\tan 3x = 0$$
The solutions are
$$\cos(2x+\frac {\pi}{2}) = 0 \space \vee \tan 3x = 0$$
Therefore
$$\cos(2x+\frac {\pi}{2}) = 0 \space \rightarrow 2x + \frac {\pi}{2} = \frac {\pi}{2} + k\pi \rightarrow x = \frac {\pi}{2} k$$
and 
$$\tan3x = 0 \rightarrow 3x = k\pi \rightarrow x = \frac {\pi}{3}k$$
The solutions should be
$$x = \frac {\pi}{2}k \vee x=\frac{\pi}{3}k$$
The solution suggested by my book is this one
$$x = \frac {\pi}{3}k$$
Why is my solution different? Any hints?

The problem is that while $$\lim_{x \to \frac{\pi}{2}} \left[ \cos\left( 2x+\frac{\pi}{2} \right) \right]=0$$
is true, $$\lim_{x \to \frac{\pi}{2}} \left[ \tan\left(3x\right) \right]=\infty$$
so we cannot not just say that the product is equal to zero. We can take the limit of the product, 
$$\lim_{x \to \frac{\pi}{2}} \left[ \cos\left( 2x+\frac{\pi}{2} \right) \tan \left(3x\right)\right]=-\frac{2}{3}\not=0$$
Thus only $\frac{\pi}{3}k$ works. 
