We have 0 = f ′(t) = −18(2 sin(t) − 1)(sin(t) + 1) when 0 = 2 sin(t) − 1 or 0 = sin(t) + 1, which means that the only critical number in the interval 0, 𝜋2 is at t =

To find the critical number in the interval 0, 𝜋/2, we need to solve the equations 0 = 2 sin(t) - 1 and 0 = sin(t) + 1.

For the first equation, 0 = 2 sin(t) - 1, we add 1 to both sides to get 1 = 2 sin(t). Then, we divide both sides by 2 to get sin(t) = 1/2. The solution to this in the interval 0, 𝜋/2 is t = 𝜋/6.

For the second equation, 0 = sin(t) + 1, we subtract 1 from both sides to get -1 = sin(t). However, in the interval 0, 𝜋/2, sin(t) is always greater than or equal to 0, so there are no solutions to this equation in the given interval.

Therefore, the only critical number in the interval 0, 𝜋/2 is at t = 𝜋/6.