For the function f(x)=cos(x/6), find all interval(s) where f(x)>0 on the interval (0,24π).
(3π,9π)∪(15π,21π)
(0,3π)∪(6π,9π)∪(12π,15π)∪(18π,21π)
(6π,12π)∪(18π,24π)
(0,3π)∪(9π,15π)∪(21π,24π)
(0,6π)∪(12π,18π)
is it B?
thanks
I do not agree with you. x/6 is zero to pi/2 and 3/2 pi to 2 pi, around and around
if x/6 = 3/2 pi
x = 9 pi
this is the only one that starts that way:
(0,3π)∪(9π,15π)∪(21π,24π)
the period is 12π, and f(x) > 0 in the 1st and 4th quarters of each period. so it looks like D to me.
Love you guys :)
To find the interval(s) where f(x) > 0 on the interval (0, 24π), you need to determine where the function is positive.
Start by solving the inequality f(x) > 0 for the function f(x) = cos(x/6).
cos(x/6) > 0
To find the intervals where cos(x/6) > 0, consider the unit circle and its quadrants.
In the unit circle, cos(x/6) > 0 when x/6 lies within the first and fourth quadrants.
The first quadrant is from 0 to π/2 (0 to 6π/6) and the fourth quadrant is from 2π to 5π/2 (12π/6 to 15π/6).
Thus, the solution for the inequality cos(x/6) > 0 is:
(0, 6π/6) U (12π/6, 15π/6)
Now, simplify the solution by converting it back to the original interval (0, 24π).
(0, 6π) U (12π, 15π)
Therefore, the answer is (0, 6π) U (12π, 15π), which corresponds to option (E).