Solving trigonometric inequalities

[LovePhys]

Hello,

I am struggling with solving trigonometric inequalities. For example, solve: cos(\frac{\pi t}{3}) < \frac{1}{2}, 0<t<50
I wonder if one of these solutions is true:
1/ \frac{\pi}{3} + k2\pi < \frac{\pi t}{3} < \frac{5\pi}{3} + k2\pi, k \in Z
2/ \frac{\pi}{3} + 6k < \frac{\pi t}{3} < \frac{5\pi}{3} + 6k, k \in Z (the period of cos(\frac{\pi t}{3}) 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 \frac{2\pi}{3}. Obviously, it'll repeat with the period of 2\pi 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 cos(\frac{\pi t}{3}).
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

[DiracRules]

If you leave them this way, both of them are not correct.

Think of which is the variable you need to find, and the limits on it posed by the problem.
The problem asks to find t.
So, try to put the solutions you wrote in the form ...<t<...
Then, you must find the intervals that satisfy both the trigonometric inequality and the condition on t.

What does this mean?

[LovePhys]

@ DriracRules: Thanks for your response. I have been thinking about this problem, yet I haven't figured it out. If you know the answer, can you please help me? Thank you very much.

[HallsofIvy]

First, note that cos(\pi/3)= 1/2 and cosine is decreasing between 0 and \pi so, immediately, cos(x)< 1/2 for \pi/3< x\le \pi which, for this problem, gives \pi/2< \pi t/3\le \pi. Solve for t and then use the periodicity of cosine to extend to values of t< 50.