- #1
LovePhys
- 57
- 0
Hello,
I am struggling with solving trigonometric inequalities. For example, solve: [itex]cos(\frac{\pi t}{3}) < \frac{1}{2}[/itex], [itex] 0<t<50 [/itex]
I wonder if one of these solutions is true:
1/ [itex] \frac{\pi}{3} + k2\pi < \frac{\pi t}{3} < \frac{5\pi}{3} + k2\pi, k \in Z [/itex]
2/ [itex] \frac{\pi}{3} + 6k < \frac{\pi t}{3} < \frac{5\pi}{3} + 6k, k \in Z [/itex] (the period of [itex]cos(\frac{\pi t}{3})[/itex] is 6)
I checked both of them and it seemed that the first solution is correct. However, personally, I think both of them are correct:
1/ The first solution: For example, we got the solution [itex] \frac{2\pi}{3}[/itex]. Obviously, it'll repeat with the period of [itex]2\pi[/itex] on the unit circle.
2/ The second solution: If we got one solution, it'll repeat with the period of 6 on the graph of [itex]cos(\frac{\pi t}{3})[/itex].
I have been struggling with this problem for a long time, yet I cannot figure it out.
Hopefully I can be given a little help.
Thanks a bunch everyone!
Huyen Nguyen
I am struggling with solving trigonometric inequalities. For example, solve: [itex]cos(\frac{\pi t}{3}) < \frac{1}{2}[/itex], [itex] 0<t<50 [/itex]
I wonder if one of these solutions is true:
1/ [itex] \frac{\pi}{3} + k2\pi < \frac{\pi t}{3} < \frac{5\pi}{3} + k2\pi, k \in Z [/itex]
2/ [itex] \frac{\pi}{3} + 6k < \frac{\pi t}{3} < \frac{5\pi}{3} + 6k, k \in Z [/itex] (the period of [itex]cos(\frac{\pi t}{3})[/itex] is 6)
I checked both of them and it seemed that the first solution is correct. However, personally, I think both of them are correct:
1/ The first solution: For example, we got the solution [itex] \frac{2\pi}{3}[/itex]. Obviously, it'll repeat with the period of [itex]2\pi[/itex] on the unit circle.
2/ The second solution: If we got one solution, it'll repeat with the period of 6 on the graph of [itex]cos(\frac{\pi t}{3})[/itex].
I have been struggling with this problem for a long time, yet I cannot figure it out.
Hopefully I can be given a little help.
Thanks a bunch everyone!
Huyen Nguyen