Simple trigonometry question I am just wondering how can you get from 
$\cos(\pi t)=1/2 $    or $\cos(\pi t)=-1$
for $0<t<4$
to 
t = 1/3, 1
t = 5/3, 7/3, 11/3 ,3 
I got 
$(\pi t) = \pi/3 +2k\pi $  and 
$(\pi t) = 5\pi/3 + 2k\pi $
$t = 1/3 + 2k $   and $t = 5/3 +2k$
but i couldn't quite get t = 3, and t =1.... i'm not sure where i got it wrong... so could someone please explain this clearly to me?
 A: There are two values at which $\cos(\pi t)$ is a solution. You've made a dent in finding two of the four solutions to $$\cos(\pi t) = \frac 12, \;\;\text{for}\;\;t \in (0, 4).$$


*

*There are four values at which $t$ gives a solution; they occur when $$0 \lt t \lt 4 \iff 0 \lt \theta \lt  4\pi$$

*You found $t = \dfrac 13,\;\dfrac 53$, but also note that $t = \dfrac 73, \dfrac {11}{3} \implies \theta = \frac {7\pi}3, \theta = \frac{11}{\pi} \lt 4\pi = \pi t$



But you also found that $\cos(\pi t) = -1$ is also a solution. This happens when $t$ is an odd integer, so that $\pi t$ is an odd multiple of $\pi$:
$$\cos(\pi t) = -1 \implies \;\;t = 1, \;\;\text{or}\;\; t = 3, \quad0\lt t \lt 4$$
A: $t = 1, 3$ look like the solution to $\cos(\pi t) = -1$, since $\cos(k\pi) =  -1$ for odd $k \in \mathbb{Z}$. Since you've restricted $0 < t < 4$, these are the two solutions.
A: You solved for $t$ s.t. $\cos\pi t=1/2$, but you also must solve for $\cos\pi t=-1$:$$\pi t=\pi+2\pi k\\\implies t=1+2k$$For $t\in(0,4)$ we find $t=1,3$ are solutions.
