Circular motion trig We have $x_P = -2 + 4 \cos (-\pi t)$ and $y_P = 1 + 4 \sin ( - \pi t)$ with $t$ in seconds. 
We have to find the coordinates of the intersection with the y-axis. So I use trig and I eventually end up with: 
$ t = -\dfrac{1}{3} + k \cdot 2$ and $ t = \dfrac{1}{3} + k \cdot 2$
The correction model does this too, and then they just say 'the corrosponding points are ...' and give the 2 points. But what I don't understand is how this can be correct. If we fill in for k, we get 6 points within $[0, 2\pi]$. This would imply 6 intersections, right? What am I missing here? 
p.s. - $t$ would be 1/3, 7/3, 10/3, 5/3, 11/3, 17/3  

$$\text{If }t=2k+ \frac13,$$ 
$$\sin(-\pi t)=-\sin \pi t=-\sin\left(2k\pi+\frac\pi3\right)=-\sin\left(\frac\pi3\right)=-\frac{\sqrt3}2$$
$$\text{If }t=2k- \frac13,$$ 
$$\sin(-\pi t)=-\sin \pi t=-\sin\left(2k\pi-\frac\pi3\right)=-\sin\left(-\frac\pi3\right)=\sin\left(\frac\pi3\right)=+\frac{\sqrt3}2$$
So, the two coordinates of the intersection with the y-axis are $\left(0,1+4\cdot\pm \frac{\sqrt3}2 \right)=\left(0,1\pm2\sqrt3\right)$

Alternatively, as $-1\le \cos(-\pi t)\le 1\implies-3\le x_p\le 2$
If $x_p$ assumes any value in the above range $=c$(say), 
$-2+ 4\cos(-\pi t)=c\implies \cos(-\pi t)=\frac{c+2}4$
$$\implies \sin (-\pi t)=\pm\frac{\sqrt{\left(4^2-(c+2)^2\right)}}4=\pm\frac{\sqrt{\left(12-c^2-4c\right)}}4$$
$$\implies y_p=1+4\left( \pm\frac{\sqrt{\left(12-c^2-4c\right)}}4\right)=1\pm \sqrt{12-c^2-4c}$$
So, the intersections will be $\left(c,1\pm \sqrt{12-c^2-4c}\right)$
