Undergrad Why is this SHM the way it is?

[Vivek98phyboy]

I know four different forms in which an SHM can be represented after solving the differential and taking the superposition
acos(wt+Ø)
asin(wt+Ø)
acos(wt-Ø)
asin(wt-Ø)
where a- amplitude
In the above image they took B as negative in order to arrive at acos(wt+e). If i already knew i wanted acos(wt+e) or acos(wt-e) then i would have decided how to modify B in order to get my answer, but if I'm deriving it for the first time then how would i know whether to take B as +ve or -ve
Why is it necessary to take B as negative, can't they just take it as +ve and end up with cos(wt-Ø). How is the process justified here?

 

[PeroK]

View attachment 253960
I know four different forms in which an SHM can be represented after solving the differential and taking the superposition
acos(wt+Ø)
asin(wt+Ø)
acos(wt-Ø)
asin(wt-Ø)
where a- amplitude
In the above image they took B as negative in order to arrive at acos(wt+e). If i already knew i wanted acos(wt+e) or acos(wt-e) then i would have decided how to modify B in order to get my answer, but if I'm deriving it for the first time then how would i know whether to take B as +ve or -ve
Why is it necessary to take B as negative, can't they just take it as +ve and end up with cos(wt-Ø). How is the process justified here?

It's all justified by the trig identities leading to equation 22.

Note they are not taking $B$ as negative. $B$ can be +ve or -ve here. They are defining $\epsilon$ so that $\sin \epsilon = \frac{-B}{\sqrt{A^2 + B^2}}$. The -ve sign is needed do that the required trig identity holds.