A (probably) easy M3 question

Paper in question (Q. 5b)
Mark scheme

In question 5.b, why are they just using T=mgSin(theta) to find out the tension when Sin(theta)=1/3 but they're using the circular motion formula when the Sin(theta)=1/2? Why don't you use the circular motion formula in the first case (or why don't you use the T=mgSin(theta) relationship in the second case)? Thanks.

Paper in question (Q. 5b)
Mark scheme

In question 5.b, why are they just using T=mgSin(theta) to find out the tension when Sin(theta)=1/3 but they're using the circular motion formula when the Sin(theta)=1/2? Why don't you use the circular motion formula in the first case (or why don't you use the T=mgSin(theta) relationship in the second case)? Thanks.

The general formula for this question is $T=mv^2/r+mg\sin(\theta)$

When $\sin(\theta)=1/3$, v=0 and the first term is zero, hence....

When $\theta=\pi /2$, v is not zero, and you have to account for it.

I did terrible at M3, but here's an attempt at an answer:

They are using the same formula in both cases. For sin(theta)=1/3, V=0 so the circular motion part of T is 0. For theta=pi/2, sin(pi/2)=1 so you could write this as T=(mv^2)/3a+mgsin(pi/2).