solving for θ

[xxMsJojoxx]

Why does this sin θ = 0 reduce to the below?
θ = nπ

I understand that it can also be , as we are trying to isolate θ .

But I don't get how we get θ = nπ.

 

[Dr.Peterson]

Why does this sin θ = 0 reduce to the below?
θ = nπ

I understand that it can also be View attachment 22639, as we are trying to isolate θ .

But I don't get how we get θ = nπ.

Because sin(0) = 0, and sin(π) = 0, and sin(θ + 2nπ) = sin(θ).

That is, there are two values of θ in each cycle whose sine is 0. The inverse sine gives only one of these infinitely many solutions, namely 0.

An easy way to think about this is to look at the graph of sine; another is to look at the unit circle.