3

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?

apnorton
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Moses
  • 65

3 Answers3

3

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 $\pi$:

$$\cos(\pi t) = -1 \implies \;\;t = 1, \;\;\text{or}\;\; t = 3, \quad0\lt t \lt 4$$

amWhy
  • 209,954
  • Hi thank you very much for your response, I got 1/3, 5/3 and 7/3, 11/3

    but i am just wondering why 13/3 and 17/3 are not the answers, is it because they are not within the range of 0<theta<4*pi ?

     – Moses May 29 '13 at 00:31
  • The way i do it is just simplifying the expression: cos(pit) = 1/2 => pit = pi/3 +2kpi or pit = 5pi/3 + 2kpi i can get 7/3 and 11/3 if i substitute 1 into the value of k, but what if i substitute 2, then i get 7/3 and 11/3 – Moses May 29 '13 at 00:31
  • With $t = 13/3$, we have $\theta = \frac{13\pi}3 > 4\pi$, same with $t = 17/3 \implies \theta = \frac{17\pi}{3} > 4\pi$ – amWhy May 29 '13 at 00:34
  • okay thank you very much, i didn't check that 13/3 * pi was greater than 4pi – Moses May 29 '13 at 00:37
  • @amWhy: Gets another TU! +1 – Amzoti May 29 '13 at 00:38
  • You're welcome! – amWhy May 29 '13 at 00:38
2

$t = 1, 3$ 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.

Alex Wertheim
  • 20,278
0

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.

obataku
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